Free Math Worksheets

- What does the product rule say about the derivative of a product of two functions? - How can you multiply the two polynomial factors systematically? - Which derivative rules apply after the expression is written as a polynomial? - Compare the two results. They should be identical.

1. Let \(u(x) = x^2 - 2x + 3\) and \(v(x) = 2x - 5\). Then \(u'(x) = 2x - 2\) and \(v'(x) = 2\). By the product rule, \(f'(x) = u'(x)v(x) + u(x)v'(x)\) \(= (2x - 2)(2x - 5) + 2(x^2 - 2x + 3)\) \(= 6x^2 - 18x + 16\). 2. First expand: \(f(x) = 2x^3 - 9x^2 + 16x - 15\). Differentiating term by term gives \(f'(x) = 6x^2 - 18x + 16\).

\(f'(x) = 6x^2 - 18x + 16\)

- What two functions are being multiplied? - What is the derivative of cosine? Pay close attention to the sign. - Differentiate each factor before applying the product rule. - Use parentheses carefully when substituting into the product rule.

1. Let \(u(x) = x^2 - 5x + 3\) and \(v(x) = \cos(x)\). 2. Then \(u'(x) = 2x - 5\) and \(v'(x) = -\sin(x)\). 3. Apply the product rule: \(f'(x) = u'(x)v(x) + u(x)v'(x)\) \(= (2x - 5)\cos(x) - (x^2 - 5x + 3)\sin(x)\).

\(f'(x) = (2x - 5)\cos(x) - (x^2 - 5x + 3)\sin(x)\)

- What two functions are being multiplied? - Find the derivative of each factor separately. - What is the structure of the product rule? - Is there a common factor in the two terms of the derivative?

1. Let \(u(x) = x^2 - 5\) and \(v(x) = e^x\). 2. Differentiate the factors: \(u'(x) = 2x\) and \(v'(x) = e^x\). 3. Apply the product rule: \(f'(x) = u'(x)v(x) + u(x)v'(x)\) \(= 2xe^x + (x^2 - 5)e^x\). 4. Factor out \(e^x\): \(f'(x) = (x^2 + 2x - 5)e^x\).

\(f'(x) = (x^2 + 2x - 5)e^x\)

- Recall the product rule for \((uv)'(x)\). - Differentiate \(u\) and \(v\) separately; the chain rule can be used for \(v\). - Compute both \((uv)'(x)\) and \(u'(x)v'(x)\). - Compare the expressions over the full domain \(x \ne 1\).

1. Differentiate the factors: \(u'(x) = 1\) and, by the chain rule, \(v'(x) = \frac{1}{(1-x)^2}\). 2. Use the product rule: \((uv)'(x) = u'(x)v(x) + u(x)v'(x)\) \(= \frac{1}{1-x} + \frac{x}{(1-x)^2}\) \(= \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}\). 3. The product of the separate derivatives is \(u'(x)v'(x) = \frac{1}{(1-x)^2}\). Thus the two expressions are equal for every \(x \ne 1\), so the claim is true for this specific pair of functions, even though it is not a general derivative rule.

The claim is true for these functions because both sides equal \(\frac{1}{(1-x)^2}\) for every \(x \ne 1\).

- A horizontal tangent has derivative value \(0\). - Use the product rule. - Factor out the common exponential term. - The exponential function is never zero.

1. Apply the product rule: \(f_k'(x)=(kx+k-1)e^x\). 2. A horizontal tangent at \(x=3\) requires \(f_k'(3)=0\). 3. Thus, \((4k-1)e^3=0\). 4. Since \(e^3\neq0\), solve \(4k-1=0\) to obtain \(k=\frac{1}{4}\).

\(k=\frac{1}{4}\)

- Identify the two factors in the function. - Use the chain rule when differentiating \(e^{2x}\). - Factor out the common exponential term at the end.

1. Let \(u(x) = 3x^2 - 5x\) and \(v(x) = e^{2x}\). 2. Then \(u'(x) = 6x - 5\) and, by the chain rule, \(v'(x) = 2e^{2x}\). 3. Apply the product rule: \(f'(x) = (6x-5)e^{2x} + (3x^2-5x)(2e^{2x})\). 4. Factor out \(e^{2x}\) and combine like terms: \(f'(x) = (6x^2 - 4x - 5)e^{2x}\).

\(f'(x) = (6x^2 - 4x - 5)e^{2x}\)

- Rewrite the square root as a power if that makes it easier to differentiate. - Identify and differentiate the two factors separately. - What is the derivative of sine? - Combine the derivatives using the product rule.

1. Let \(u(x)=\sqrt{x}+2\) and \(v(x)=\sin(x)\). 2. Then \(u'(x)=\frac{1}{2\sqrt{x}}\) and \(v'(x)=\cos(x)\). 3. Apply the product rule: \(f'(x)=\frac{\sin(x)}{2\sqrt{x}}+(\sqrt{x}+2)\cos(x)\).

\(f'(x)=\frac{\sin(x)}{2\sqrt{x}}+(\sqrt{x}+2)\cos(x)\), for \(x>0\)

- Substitute the coordinates of the given point. - Use the product and chain rules to differentiate. - Evaluate the derivative at \(x=0\) and solve for \(a\).

1. Substitute the point: \((2+a)e^{-2}=5e^{-2}\). Dividing by \(e^{-2}\) gives \(a=3\). 2. By the product rule, \(g_a'(x)=(1-x-a)e^{-x}\). Thus, \(g_a'(0)=1-a\). Setting \(1-a=-2\) gives \(a=3\).

1. \(a=3\) 2. \(a=3\)

- First find the point on the graph at \(x = 0\). - Which differentiation rule applies to a product of two functions? - What does the derivative at a point tell you geometrically? - How are the slopes of perpendicular lines related? - Use point-slope form for each line.

1. Find the point on the graph: \(f(0) = (2 - 0)e^0 = 2\), so the point of tangency is \(P(0, 2)\). 2. Differentiate using the product rule: \(f'(x) = (-1)e^x + (2 - x)e^x = (1 - x)e^x\). 3. The tangent slope is \(m_t = f'(0) = 1\). Using point-slope form, \(y - 2 = 1(x - 0)\), so \(t: y = x + 2\). 4. The normal slope is the negative reciprocal of \(1\), so \(m_n = -1\). Thus \(y - 2 = -1(x - 0)\), and \(n: y = -x + 2\).

Tangent line: \(t: y = x + 2\) Normal line: \(n: y = -x + 2\)

- Identify the two factors and differentiate each one. - Recall the power rule for derivatives. - When expanding, pay close attention to signs and exponents. - Think about why the product rule can be more efficient when the factors are more complicated.

1. Using the product rule, let \(u(x) = 2x^2 - 4x\) and \(v(x) = x^2 + 1\). Then \(u'(x) = 4x - 4\) and \(v'(x) = 2x\), so \(g'(x) = (4x - 4)(x^2 + 1) + (2x^2 - 4x)(2x)\) \(= 8x^3 - 12x^2 + 4x - 4\). 2. Expanding first gives \(g(x) = 2x^4 - 4x^3 + 2x^2 - 4x\). Differentiating term by term gives \(g'(x) = 8x^3 - 12x^2 + 4x - 4\), which matches the product-rule result.

\(g'(x) = 8x^3 - 12x^2 + 4x - 4\)

- One factor is a sum. How do you differentiate that sum? - Use the product rule as the main structure of your work. - After differentiating, try grouping the terms containing \(\sin(x)\) and \(\cos(x)\). - Watch the signs when differentiating sine and cosine.

1. Let \(u(x) = x^2\) and \(v(x) = \sin(x) + \cos(x)\). 2. Then \(u'(x) = 2x\) and \(v'(x) = \cos(x) - \sin(x)\). 3. Apply the product rule: \(g'(x) = 2x(\sin(x) + \cos(x)) + x^2(\cos(x) - \sin(x))\). 4. Group the sine and cosine terms: \(g'(x) = (2x - x^2)\sin(x) + (2x + x^2)\cos(x)\).

\(g'(x) = (2x - x^2)\sin(x) + (2x + x^2)\cos(x)\)

- Recall the product rule for a function of the form \(u(x)v(x)\). - Check the signs carefully when differentiating trigonometric functions. - Compare each proposed expression with the structure of the product rule. - Ask whether any work simply multiplied the derivatives of the two factors.

1. In A, the derivatives of the two factors were multiplied instead of using the product rule. The correct derivative is \(f'(x) = 2x\sin(x) + x^2\cos(x)\). 2. In B, subtraction was used instead of addition in the product rule. The correct derivative is \(g'(x) = e^x(x^3 - 2) + e^x(3x^2) = e^x(x^3 + 3x^2 - 2)\). 3. In C, the derivative of \(\cos(x)\) was given the wrong sign. Since \(\frac{d}{dx}\cos(x) = -\sin(x)\), \(h'(x) = -(5x^2 + 2)\sin(x) + 10x\cos(x)\).

A) Error: The derivatives of the factors were multiplied instead of applying the product rule. Correct: \(f'(x) = 2x\sin(x) + x^2\cos(x)\). B) Error: A minus sign was used where the product rule requires a plus sign. Correct: \(g'(x) = e^x(x^3 + 3x^2 - 2)\). C) Error: The derivative of cosine has a negative sign. Correct: \(h'(x) = -(5x^2 + 2)\sin(x) + 10x\cos(x)\).

- Use the product rule, then factor out \(e^x\). - What must the derivative equal when a tangent line is horizontal? - Can \(e^x\) ever equal zero? - Substitute each resulting \(x\)-value into the original function to find the full coordinates.

1. Let \(u(x) = x^2 - 8\) and \(v(x) = e^x\). Then \(u'(x) = 2x\) and \(v'(x) = e^x\). 2. By the product rule, \(f'(x) = 2xe^x + (x^2 - 8)e^x = (x^2 + 2x - 8)e^x\). 3. A horizontal tangent occurs where \(f'(x) = 0\). Since \(e^x > 0\) for every real \(x\), solve \(x^2 + 2x - 8 = (x - 2)(x + 4) = 0\). Thus \(x = 2\) or \(x = -4\). 4. Evaluate the original function: \(f(2) = -4e^2\) and \(f(-4) = 8e^{-4}\). The points are \((2, -4e^2)\) and \((-4, 8e^{-4})\).

a) \(f'(x) = (x^2 + 2x - 8)e^x\) b) \((2, -4e^2)\) and \((-4, 8e^{-4})\)

- Recall the derivative of \(\ln(x)\). - Treat \(x\ln(x)\) as a product of two functions. - Find where the first derivative is zero, then classify that critical point.

1. Write \(g(x)=u(x)v(x)\), where \(u(x)=x\) and \(v(x)=\ln(x)\). Then \(u'(x)=1\) and \(v'(x)=\frac{1}{x}\). 2. Apply the product rule: \(g'(x)=u'(x)v(x)+u(x)v'(x)=\ln(x)+1\). 3. Set the first derivative equal to zero: \(\ln(x)+1=0\), so \(\ln(x)=-1\) and \(x=e^{-1}=\frac{1}{e}\). 4. Differentiate again: \(g''(x)=\frac{1}{x}\). Since \(g''\left(\frac{1}{e}\right)=e>0\), the function has a local minimum at \(x=\frac{1}{e}\). 5. Its y-coordinate is \(g\left(\frac{1}{e}\right)=\frac{1}{e}\ln\left(\frac{1}{e}\right)=-\frac{1}{e}\).

The function has a local minimum at \(\left(\frac{1}{e},-\frac{1}{e}\right)\). The product rule gives \(g'(x)=\ln(x)+1\), and the critical-point equation \(g'(x)=0\) gives \(x=\frac{1}{e}\).

- Treat the function as the sum of a polynomial and a product. - Differentiate each part separately. - Use the product rule on the term containing \(\sin(x)\). - Combine all differentiated terms at the end.

1. Differentiate the polynomial part: \(\frac{d}{dx}(3x^2 - 7x) = 6x - 7\). 2. For \((5x - x^3)\sin(x)\), let \(u(x) = 5x - x^3\) and \(v(x) = \sin(x)\). Then \(u'(x) = 5 - 3x^2\) and \(v'(x) = \cos(x)\). 3. Apply the product rule to that term: \(\frac{d}{dx}[(5x - x^3)\sin(x)] = (5 - 3x^2)\sin(x) + (5x - x^3)\cos(x)\). 4. Combine the derivatives: \(g'(x) = 6x - 7 + (5 - 3x^2)\sin(x) + (5x - x^3)\cos(x)\).

\(g'(x) = 6x - 7 + (5 - 3x^2)\sin(x) + (5x - x^3)\cos(x)\)

- What do the tangent line and its slope tell you about \(v(2)\) and \(v'(2)\)? - Which derivative rule applies to a product of two functions? - How is the slope of a tangent line related to a derivative value? - Find a general expression for \(f'(x)\) before substituting \(x = 2\).

1. Because \(t\) is tangent to the graph of \(v\) at \(x = 2\), \(v(2) = t(2) = 3 \cdot 2 + 1 = 7\) and \(v'(2) = 3\), the slope of the tangent line. 2. Apply the product rule: \(f'(x) = 2xv(x) + (x^2 + 4)v'(x)\). 3. Substitute the known values: \(f'(2) = 2 \cdot 2 \cdot 7 + (2^2 + 4) \cdot 3 = 28 + 24 = 52\).

The slope is \(52\).

- Which rule applies when a function is written as a product? - List the known function and derivative values at \(x = 0\). - What is the derivative of \(e^x\)? - How does a derivative value relate to the slope of a tangent line?

1. Apply the product rule: \(h'(x) = e^xg(x) + (e^x + 2)g'(x)\). 2. Substitute \(x = 0\), using \(e^0 = 1\), \(g(0) = 4\), and \(g'(0) = -1\): \(h'(0) = 1 \cdot 4 + (1 + 2) \cdot (-1) = 4 - 3 = 1\).

The slope is \(1\).

- Recall the product rule for \((fg)'(x)\). - Differentiate \(f\) and \(g\) separately. - Factor the common exponential term from both expressions. - Compare the coefficients of the common exponential factor.

1. Differentiate the two functions: \(f'(x) = 2e^{2x}\) and \(g'(x) = ke^{kx}\). 2. Use the product rule: \((fg)'(x) = f'(x)g(x) + f(x)g'(x)\) \(= 2e^{2x}e^{kx} + ke^{2x}e^{kx} = (2+k)e^{(2+k)x}\). 3. The product of the separate derivatives is \(f'(x)g'(x) = 2ke^{(2+k)x}\). 4. Set the expressions equal. Since \(e^{(2+k)x}\) is never zero, \(2+k = 2k\). Thus \(k = 2\).

\(k = 2\)

- What function value and derivative value can you read from the given point and horizontal tangent? - Use the product rule for each new function. - What must a derivative equal at a point with a horizontal tangent?

1. The given point and horizontal tangent imply \(f(1) = -2\) and \(f'(1) = 0\). 2. For \(k\), apply the product rule: \(k'(x) = 2xf(x) + x^2f'(x)\). Then \(k'(1) = 2 \cdot 1 \cdot (-2) + 1^2 \cdot 0 = -4\). Since \(k'(1) \ne 0\), the graph of \(k\) does not have a horizontal tangent at \(x = 1\). 3. For \(m\), apply the product rule: \(m'(x) = f'(x)e^{x-1} + (f(x) + 2)e^{x-1}\). Then \(m'(1) = 0e^0 + (-2 + 2)e^0 = 0\). Therefore, the graph of \(m\) does have a horizontal tangent at \(x = 1\).

The graph of \(k\) does not have a horizontal tangent at \(x = 1\) because \(k'(1) = -4\). The graph of \(m\) does have a horizontal tangent at \(x = 1\) because \(m'(1) = 0\).

- List all known function and derivative values at \(x = 0\). - What condition must the derivative satisfy for a horizontal tangent? - Apply the product rule and substitute the known values.

1. The given information is \(u(0) = 2\), \(u'(0) = -4\), and \(v(0) = 1\). 2. A horizontal tangent at \(x = 0\) requires \(h'(0) = 0\). 3. By the product rule, \(h'(x) = u'(x)v(x) + u(x)v'(x)\). At \(x = 0\), \(-4 \cdot 1 + 2v'(0) = 0\). 4. Solving gives \(2v'(0) = 4\), so \(v'(0) = 2\).

\(v'(0) = 2\)

- Extend the product rule by differentiating one factor at a time while leaving the other two unchanged. - What happens to terms in the derivative when you substitute a zero of one factor? - Examine the factored derivative before expanding it.

1. Differentiate one factor at a time: \(p'(x) = (x - 4)(x - 6) + (x - 2)(x - 6) + (x - 2)(x - 4)\). Expanding gives \(p'(x) = 3x^2 - 24x + 44\). 2. Evaluate the derivative: \(p'(2) = 8\), \(p'(4) = -4\), and \(p'(6) = 8\). 3. At each zero, the two derivative terms that still contain the zero factor vanish. The remaining term equals the product of the other two factors: \((2 - 4) \cdot (2 - 6) = 8\), \((4 - 2) \cdot (4 - 6) = -4\), \((6 - 2) \cdot (6 - 4) = 8\). Thus, at a zero of a product of distinct linear factors, the derivative equals the product of the values of all the other factors at that zero.

1. \(p'(x) = (x - 4)(x - 6) + (x - 2)(x - 6) + (x - 2)(x - 4) = 3x^2 - 24x + 44\) 2. \(p'(2) = 8\), \(p'(4) = -4\), and \(p'(6) = 8\) 3. At each zero, the slope equals the product of the values of the other two linear factors.

- At a common point, what equation must the two function values satisfy? - Why may you divide by \(f(x)\)? - Use the product rule to find \(g'(x)\). - How do the slopes distinguish a tangency from a crossing?

1. At a common point, \(g(x) = f(x)\), so \((x^2 - 8)f(x) = f(x)\). Since \(f(x) \ne 0\), divide by \(f(x)\): \(x^2 - 8 = 1\). Thus \(x^2 = 9\), so \(x = -3\) or \(x = 3\). 2. By the product rule, \(g'(x) = 2xf(x) + (x^2 - 8)f'(x)\). At \(x=3\), \(g'(3) = 6f(3) + f'(3)\). Since \(f(3) \ne 0\), \(g'(3) \ne f'(3)\). At \(x=-3\), \(g'(-3) = -6f(-3) + f'(-3)\). Since \(f(-3) \ne 0\), \(g'(-3) \ne f'(-3)\). The graphs have equal function values but different slopes at both common points, so they cross rather than being tangent.

1. \(x=-3\) and \(x=3\) 2. At both common points, \(g'(x) \ne f'(x)\), so the graphs have different tangent slopes and are not tangent to each other.

- Use the given point to write a function-value equation. - The derivative at the point gives the tangent slope. - Parallel lines have equal slopes. - Apply the product rule.

1. The point condition gives \(f_{a,b}(0)=b=-3\). 2. Use the product rule: \(f_{a,b}'(x)=(ax+a+b)e^x\). 3. A line parallel to \(y=2x+7\) has slope \(2\), so \(f_{a,b}'(0)=2\). 4. Substitute \(b=-3\): \(a-3=2\), so \(a=5\).

\(a=5\) and \(b=-3\)

- Identify the two factors in the function. - Use the product rule to differentiate. - A product is zero when at least one factor is zero. - Can \(e^x\) ever equal zero? - What must the derivative equal at a horizontal tangent?

1. Apply the product rule: \(f'(x) = (4x - 8)e^x + (2x^2 - 8x)e^x\) \(= (2x^2 - 4x - 8)e^x\). 2. To find the zeros of \(f\), solve \((2x^2 - 8x)e^x = 0\). Since \(e^x > 0\), \(2x(x-4) = 0\), so \(x = 0\) or \(x = 4\). 3. Horizontal tangents occur where \(f'(x)=0\). Again, \(e^x > 0\), so solve \(2x^2 - 4x - 8 = 0\), or \(x^2 - 2x - 4 = 0\). The quadratic formula gives \(x = 1 \pm \sqrt{5}\).

Derivative: \(f'(x) = (2x^2 - 4x - 8)e^x\) Zeros of \(f\): \(x = 0\) and \(x = 4\) Horizontal tangents: \(x = 1 - \sqrt{5}\) and \(x = 1 + \sqrt{5}\)

- Use the product rule for the polynomial and exponential factors. - Factor common terms when finding zeros. - Which factor can never equal zero? - After factoring out \(x\) from the cubic expression, solve the remaining quadratic.

1. Apply the product rule: \(f'(x) = (3x^2 - 2x)e^x + (x^3 - x^2)e^x\) \(= (x^3 + 2x^2 - 2x)e^x\). 2. To find the zeros of \(f\), solve \((x^3 - x^2)e^x = 0\). Since \(e^x \ne 0\), \(x^2(x-1)=0\), so \(x=0\) or \(x=1\). 3. To find the zeros of \(f'\), solve \(x(x^2 + 2x - 2)e^x = 0\). Thus \(x=0\), or \(x^2 + 2x - 2=0\). The quadratic formula gives \(x=-1 \pm \sqrt{3}\).

Derivative: \(f'(x) = (x^3 + 2x^2 - 2x)e^x\) Zeros of \(f\): \(x=0\) and \(x=1\) Zeros of \(f'\): \(x=0\), \(x=-1-\sqrt{3}\), and \(x=-1+\sqrt{3}\)

- Use the product rule. - Use the chain rule when differentiating the exponential factor. - Factor out the common exponential factor after differentiating. - Combine like terms inside the remaining parentheses.

1. Let \(u(x)=4-x^2\) and \(v(x)=e^{-0.5x}\). Then \(u'(x)=-2x\) and, by the chain rule, \(v'(x)=-0.5e^{-0.5x}\). 2. Apply the product rule: \(f'(x)=(-2x)e^{-0.5x}-0.5(4-x^2)e^{-0.5x}\). 3. Factor out \(e^{-0.5x}\) and simplify: \(f'(x)=(-2x-2+0.5x^2)e^{-0.5x}\) \(=(0.5x^2-2x-2)e^{-0.5x}\).

\(f'(x)=(0.5x^2-2x-2)e^{-0.5x}\)

- A horizontal tangent has slope \(0\). - Use both the product rule and the chain rule. - An exponential factor is never zero. - Compare exponential decay or growth with the linear factor.

1. Differentiate using the product and chain rules: \(f'(x)=-2e^{0.5x}+(4-2x)(0.5)e^{0.5x}=-xe^{0.5x}\). 2. A horizontal tangent requires \(f'(x)=0\). Since \(e^{0.5x}>0\), the only solution is \(x=0\). 3. Evaluate the function: \(f(0)=4\), so the point is \((0,4)\). 4. As \(x\to-\infty\), the exponential decay dominates the linear factor, so \(f(x)\to0\). As \(x\to\infty\), the linear factor approaches \(-\infty\) while the exponential factor grows without bound, so \(f(x)\to-\infty\).

Horizontal-tangent point: \((0,4)\). \(\lim_{x\to-\infty}f(x)=0\), and \(\lim_{x\to\infty}f(x)=-\infty\).

- A tangent parallel to the x-axis has slope \(0\). - Use the product rule and factor out the exponential term. - Solve the remaining quadratic equation. - Exponential decay dominates polynomial growth as \(x\to-\infty\).

1. Differentiate using the product rule: \(f'(x)=2xe^x+(x^2-3)e^x=(x^2+2x-3)e^x\). 2. A horizontal tangent requires \(f'(x)=0\). Since \(e^x>0\), solve \(x^2+2x-3=0\), which factors as \((x-1)(x+3)=0\). Thus \(x=1\) or \(x=-3\). 3. Evaluate the function: \(f(1)=-2e\), and \(f(-3)=6e^{-3}\). The points are \((1,-2e)\) and \((-3,6e^{-3})\). 4. As \(x\to\infty\), both the polynomial factor and \(e^x\) grow without bound, so \(f(x)\to\infty\). As \(x\to-\infty\), exponential decay dominates polynomial growth, so \(f(x)\to0\).

Horizontal-tangent points: \((1,-2e)\) and \((-3,6e^{-3})\). \(\lim_{x\to\infty}f(x)=\infty\), and \(\lim_{x\to-\infty}f(x)=0\).

- The derivative gives the graph’s slope. - Combine the product and chain rules. - Evaluate the derivative at \(x=1\). - Divide by the exponential factor, which is never zero.

1. Apply the product and chain rules: \(g_a'(x)=(ax^2+2x)e^{ax}\). 2. The slope condition is \(g_a'(1)=e^a\), so \((a+2)e^a=e^a\). 3. Since \(e^a>0\), divide by \(e^a\): \(a+2=1\). 4. Therefore, \(a=-1\).

\(a=-1\)

- Use the product rule. - Substitute the given double-angle identity. - The derivative represents slope. - Recall the range of the cosine function.

1. By the product rule, \(f'(x)=\cos^2x-\sin^2x\). 2. Using the double-angle identity, \(f'(x)=\cos(2x)\). 3. The slope is the value of the derivative. Since the maximum value of cosine is \(1\), the maximum possible slope is \(1\).

a) \(f'(x)=\cos^2x-\sin^2x\) b) \(f'(x)=\cos(2x)\) c) Maximum slope: \(1\)

- What does the derivative represent at a specific point? - Differentiate the function before substituting the parameter condition. - What is \(\ln(1)\)? - Set the derivative value equal to the given slope.

1. Apply the product rule: \(g'(x) = 2x\ln(x) + (x^2+k)\frac{1}{x}\). 2. Evaluate at \(x=1\): \(g'(1) = 2 \cdot 1 \cdot \ln(1) + (1+k)\frac{1}{1}\). Since \(\ln(1)=0\), \(g'(1)=1+k\). 3. Set the derivative equal to the required slope: \(1+k=4\). Therefore, \(k=3\).

\(k=3\)

- Use the product rule. - Apply the chain rule to the exponential factor. - Factor out the common exponential after differentiating. - Evaluate the first derivative to find the slope.

1. Let \(u(x)=4x-2\) and \(v(x)=e^{0.5x}\). Then \(u'(x)=4\) and \(v'(x)=0.5e^{0.5x}\). 2. Apply the product rule: \(g'(x)=4e^{0.5x}+(4x-2)(0.5)e^{0.5x}\). Factor and simplify: \(g'(x)=[4+0.5(4x-2)]e^{0.5x}=(2x+3)e^{0.5x}\). 3. Evaluate at \(x=0\): \(g'(0)=3e^0=3\).

a) \(g'(x)=(2x+3)e^{0.5x}\) b) The slope is \(3\).

- Treat \(k\) as a constant. - A horizontal tangent has derivative value \(0\). - Substitute \(x=0\) after differentiating. - The exponential factor is never zero.

1. Apply the product and chain rules: \(f_k'(x)=ke^{-2x}-2(kx+2)e^{-2x}\). 2. Factor: \(f_k'(x)=(-2kx+k-4)e^{-2x}\). 3. A horizontal tangent at \(x=0\) requires \(f_k'(0)=0\). 4. Thus, \(k-4=0\), so \(k=4\).

a) \(f_k'(x)=(-2kx+k-4)e^{-2x}\) b) \(k=4\)

- Apply the product rule twice. - Set the function equal to its second derivative. - Divide by factors that are never zero.

1. Using the product rule, \(f_k'(x)=k(x^2+2x-1)e^x\) and \(f_k''(x)=k(x^2+4x+1)e^x\). 2. At an intersection, \(k(x^2-1)e^x=k(x^2+4x+1)e^x\). 3. Since \(k\ne0\) and \(e^x>0\), divide by \(ke^x\): \(x^2-1=x^2+4x+1\). 4. Solving gives \(x=-\frac{1}{2}\), which does not depend on \(k\).

\(x=-\frac{1}{2}\)

- Use the product rule for the outer structure. - Differentiate the radical with the chain rule. - Combine the two terms over a common denominator.

1. Let \(u(x)=2x-3\) and \(v(x)=\sqrt{x^2+5}\). Then \(u'(x)=2\) and \(v'(x)=\frac{x}{\sqrt{x^2+5}}\). 2. Apply the product rule: \(f'(x)=2\sqrt{x^2+5}+(2x-3)\frac{x}{\sqrt{x^2+5}}\). 3. Combine over a common denominator: \(f'(x)=\frac{2(x^2+5)+x(2x-3)}{\sqrt{x^2+5}}=\frac{4x^2-3x+10}{\sqrt{x^2+5}}\).

\(f'(x)=\frac{4x^2-3x+10}{\sqrt{x^2+5}}\)

- Require the square-root radicand to be nonnegative. - Use the product rule for a linear factor times a radical factor. - Apply the chain rule to the radical. - Combine the two derivative terms over a common denominator.

1. Require \(6-x\ge0\), so \(D=(-\infty,6]\). 2. Let \(u(x)=x+4\) and \(v(x)=\sqrt{6-x}\). Then \(u'(x)=1\) and \(v'(x)=-\frac{1}{2\sqrt{6-x}}\). 3. By the product rule, \(f'(x)=\sqrt{6-x}-\frac{x+4}{2\sqrt{6-x}}\). 4. Combine over a common denominator: \(f'(x)=\frac{2(6-x)-(x+4)}{2\sqrt{6-x}}=\frac{8-3x}{2\sqrt{6-x}}\), for \(x<6\).

\(D=(-\infty,6]\); \(f'(x)=\frac{8-3x}{2\sqrt{6-x}}\) for \(x<6\)

- Require the logarithm’s argument to be positive. - Apply the product rule. - Use the chain rule when differentiating \(\ln(3-x)\).

1. The logarithm requires \(3-x>0\), so \(x<3\). Therefore, \(D_f=(-\infty, 3)\). 2. Apply the product rule. Since \(\frac{d}{dx}\ln(3-x)=-\frac{1}{3-x}=\frac{1}{x-3}\), the derivative is \(f'(x)=2x\ln(3-x)+\frac{x^2}{x-3}\).

\(D_f=(-\infty, 3)\) \(f'(x)=2x\ln(3-x)+\frac{x^2}{x-3}\)

- Use the chain rule for \(\ln(g(x))\). - Identify the two factors in part b. - Apply the product rule. - Simplify the resulting fractions.

1. For part a, apply the chain rule: \(f'(x)=\frac{\frac{1}{4}}{\frac{1}{4}x+2}=\frac{1}{x+8}\), for \(x>-8\). 2. For part b, apply the product rule: \(f'(x)=2\ln x+(2x-5)\frac{1}{x}\). Therefore, \(f'(x)=2\ln x+2-\frac{5}{x}\), for \(x>0\).

a) \(f'(x)=\frac{1}{x+8}\) b) \(f'(x)=2\ln x+2-\frac{5}{x}\)

- Apply the product rule. - Differentiate cosine. - Use the chain rule for \(\ln(x^2+1)\). - Combine the two product-rule terms.

1. Apply the product rule with \(u(x)=\cos x\) and \(v(x)=\ln(x^2+1)\). 2. Their derivatives are \(u'(x)=-\sin x\) and \(v'(x)=\frac{2x}{x^2+1}\). 3. Therefore, \(f'(x)=-\sin x\ln(x^2+1)+\frac{2x\cos x}{x^2+1}\).

\(f'(x)=\frac{2x\cos x}{x^2+1}-\sin x\ln(x^2+1)\)

- Substitute the conditions for a zero directly into \(f\). - Use the product rule to find \(f'\). - At a local extremum of a differentiable function, what is true about its derivative? - Recall the two conditions that define a double zero.

1. Statement (1) is true. If \(u(x_0)=0\), then \(f(x_0)=u(x_0)\ln(x_0)=0\). 2. By the product rule, \(f'(x)=u'(x)\ln(x)+\frac{u(x)}{x}\). If \(u\) has a local extremum at \(x_0\), then \(u'(x_0)=0\), but \(f'(x_0)=\frac{u(x_0)}{x_0}\), which need not be \(0\). Statement (2) is false. 3. If \(x_0\) is a double zero of \(u\), then \(u(x_0)=0\) and \(u'(x_0)=0\). Substituting into the derivative formula gives \(f'(x_0)=0\). Statement (3) is true.

(1) True (2) False (3) True

- A logarithm requires a positive argument. - Solve the quadratic inequality in part a. - Apply the chain rule in part a. - Apply the product rule in part b.

1. For part a, require \(25-x^2>0\), which gives \(-5<x<5\). Thus, \(D_f=(-5, 5)\). Applying the chain rule, \(f'(x)=-\frac{2x}{25-x^2}\). 2. For part b, the logarithm requires \(x>0\), so \(D_g=(0, \infty)\). Applying the product rule, \(g'(x)=2x\ln x+(x^2+1)\frac{1}{x}\), so \(g'(x)=2x\ln x+x+\frac{1}{x}\).

a) \(D_f=(-5, 5)\), \(f'(x)=-\frac{2x}{25-x^2}\) b) \(D_g=(0, \infty)\), \(g'(x)=2x\ln x+x+\frac{1}{x}\)

- Identify the outermost operation in the function. - Use the chain rule for each composite factor. - What is the derivative of \(\cos(kx)\)? - Keep track of the negative sign when combining the two product-rule terms.

1. Let \(u(x) = (x^3-2)^5\) and \(v(x) = \cos(4x)\). 2. By the chain rule, \(u'(x) = 5(x^3-2)^4(3x^2) = 15x^2(x^3-2)^4\), and \(v'(x) = -4\sin(4x)\). 3. Apply the product rule: \(f'(x) = 15x^2(x^3-2)^4\cos(4x) - 4(x^3-2)^5\sin(4x)\).

\(f'(x) = 15x^2(x^3-2)^4\cos(4x) - 4(x^3-2)^5\sin(4x)\)

- Separate the function into two terms and differentiate each one. - The first term requires both the product rule and the chain rule. - Identify the inner function in \(\sin(x^2+1)\). - Keep the constant factor and the inner derivative when differentiating the power in the second term.

1. Differentiate the two terms separately. 2. For \(3x^2\sin(x^2+1)\), use the product rule and the chain rule: \(\frac{d}{dx}[3x^2\sin(x^2+1)] = 6x\sin(x^2+1) + 3x^2\cos(x^2+1)(2x)\) \(= 6x\sin(x^2+1) + 6x^3\cos(x^2+1)\). 3. For the second term, use the chain rule: \(\frac{d}{dx}\left[\frac{1}{4}(2x-5)^8\right] = \frac{1}{4}\cdot 8(2x-5)^7\cdot 2 = 4(2x-5)^7\). 4. Include the subtraction from the original function: \(f'(x) = 6x\sin(x^2+1) + 6x^3\cos(x^2+1) - 4(2x-5)^7\).

\(f'(x) = 6x\sin(x^2+1) + 6x^3\cos(x^2+1) - 4(2x-5)^7\)

- Use the product rule. - Factor out \(e^x\) when simplifying the derivative. - Compare polynomial growth with exponential growth or decay in each direction.

1. Apply the product rule: \(f'(x)=(2x-2)e^x+(x^2-2x)e^x\). 2. Combine like terms: \(f'(x)=(x^2-2)e^x\). 3. As \(x\to\infty\), both \(x^2-2x\) and \(e^x\) grow without bound and are positive, so \(f(x)\to\infty\). 4. As \(x\to-\infty\), the polynomial factor grows in magnitude while \(e^x\to0\). Exponential decay dominates polynomial growth, so \(f(x)\to0\).

a) \(f'(x)=(x^2-2)e^x\) b) \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to-\infty}f(x)=0\)

- Use \((fg)'=f'g+fg'\) for the first derivative of a product. - Differentiate each factor before multiplying the individual derivatives. - Factor the equation after moving all terms to one side. - Two functions are identical only if they agree at every input in their domain.

1. By the product rule, \(p'(x)=f'(x)g(x)+f(x)g'(x)\). 2. Since \(f'(x)=6x^2\) and \(g'(x)=20x^3\), \(p'(x)=(6x^2)(5x^4)+(2x^3)(20x^3)=30x^6+40x^6=70x^6\). 3. The product of the individual derivatives is \(f'(x)g'(x)=(6x^2)(20x^3)=120x^5\). 4. Set the expressions equal: \(70x^6=120x^5\). Factoring gives \(x^5(70x-120)=0\), so \(x=0\) or \(x=\frac{12}{7}\). 5. The expressions agree only at those inputs, not as functions, so the proposed rule is false.

a) \(p'(x)=70x^6\) b) \(f'(x)g'(x)=120x^5\) c) The values agree only at \(x=0\) and \(x=\frac{12}{7}\); therefore, \((fg)'\neq f'g'\) in general.

- Apply the product rule and chain rule. - The exponential factor cannot equal zero. - Solve the critical-point condition for \(k\), then substitute.

1. The product and chain rules give \(f_k'(x)=e^{0.5x}(0.5k-0.5x-1)\). 2. Since the exponential factor is positive, a critical point satisfies \(0.5k-0.5x-1=0\), or \(k=x+2\). 3. Substitute this relation into the family: \(y=(x+2-x)e^{0.5x}=2e^{0.5x}\). 4. The second derivative is \(f_k''(x)=e^{0.5x}(0.25k-0.25x-1)\). At a critical point, \(k=x+2\), so \(f_k''(x)=-0.5e^{0.5x}<0\). Therefore, every critical point is a local maximum, and the locus is \(y=2e^{0.5x}\).

\(y=2e^{0.5x}\)

- Find the x-value that makes the parameter term disappear. - Use the product rule to differentiate. - A tangent line is determined by a point and a slope.

1. The parameter term is zero when \((x-1)^2=0\), so \(x=1\). Then \(g_t(1)=e\), and every graph passes through \(Q(1, e)\). 2. By the product rule, \(g_t'(x)=e^x\left(2t(x-1)+t(x-1)^2+1\right)\). 3. At \(x=1\), \(g_t'(1)=e\), independent of \(t\). The line through \((1,e)\) with slope \(e\) is \(y-e=e(x-1)\), or \(y=ex\).

The common point is \(Q(1, e)\), and the common tangent line is \(y=ex\).

- The slope at a point is the value of the derivative there. - Use the product rule on the two factors. - Rewrite \(\frac{2}{x}\) using a negative exponent if helpful. - Recall the exact values of sine and cosine at \(\pi\).

1. Write the function as a product with \(u(x)=\frac{2}{x}\) and \(v(x)=\sin(x)\). 2. Then \(u'(x)=-\frac{2}{x^2}\) and \(v'(x)=\cos(x)\). 3. Apply the product rule: \(f'(x) = -\frac{2}{x^2}\sin(x) + \frac{2}{x}\cos(x)\). 4. Evaluate at \(x=\pi\): \(f'(\pi) = -\frac{2}{\pi^2}\sin(\pi) + \frac{2}{\pi}\cos(\pi)\). Since \(\sin(\pi)=0\) and \(\cos(\pi)=-1\), \(f'(\pi)=-\frac{2}{\pi}\).

The slope is \(-\frac{2}{\pi}\).

- Rewrite \(\frac{a}{x^2}\) using a negative exponent. - Use the product rule for the two factors. - Watch the sign when differentiating cosine. - Combine the two fractions using a common denominator.

1. Write \(\frac{a}{x^2}=ax^{-2}\). Let \(u(x)=ax^{-2}\) and \(v(x)=\cos(x)\). 2. Then \(u'(x)=-2ax^{-3}=-\frac{2a}{x^3}\) and \(v'(x)=-\sin(x)\). 3. Apply the product rule: \(f_a'(x)=-\frac{2a}{x^3}\cos(x)-\frac{a}{x^2}\sin(x)\). 4. Using a common denominator, \(f_a'(x)=-\frac{a(2\cos(x)+x\sin(x))}{x^3}\).

\(f_a'(x)=-\frac{a(2\cos(x)+x\sin(x))}{x^3}\)

- Use the derivative of \(u^2\), or derive it from \(u\cdot u\). - In method a), include the derivative of the expression inside the parentheses. - In method b), use the binomial expansion for a cube. - Constants have derivative zero.

1. Write \(u^3=u^2u\). By the product rule, \((u^3)'=(u^2)'u+u^2u'\). Since \((u^2)'=2uu'\), \((u^3)'=2u^2u'+u^2u'=3u^2u'\). 2. For method a), let \(u(x)=x^2+1\), so \(u'(x)=2x\). Then \(k'(x)=3(x^2+1)^2(2x)=6x(x^2+1)^2\). Expanding this result gives \(k'(x)=6x^5+12x^3+6x\). 3. For method b), first expand: \(k(x)=x^6+3x^4+3x^2+1\). Differentiating term by term gives \(k'(x)=6x^5+12x^3+6x\). Both methods produce the same derivative.

1. \((u^3)' = 3u^2u'\) 2. \(k'(x)=6x(x^2+1)^2=6x^5+12x^3+6x\)

- Differentiate each factor before applying the three-factor rule. - What does the product rule look like when one factor at a time is differentiated? - Group two factors together and treat their product as one function. - Expand the grouped product-rule expression to obtain the three terms.

1. Let \(u(x)=x^2\), \(v(x)=e^x\), and \(w(x)=\ln(x)\). Then \(u'(x)=2x\), \(v'(x)=e^x\), and \(w'(x)=\frac{1}{x}\). 2. Apply the product rule for three factors: \(f'(x)=2xe^x\ln(x)+x^2e^x\ln(x)+x^2e^x\frac{1}{x}\). 3. Factor the result: \(f'(x)=xe^x(2\ln(x)+x\ln(x)+1)\). 4. For the general rule, write \(uvw=(uv)w\). Then \(((uv)w)'=(uv)'w+(uv)w'\). Since \((uv)'=u'v+uv'\), \(((uv)w)'=(u'v+uv')w+uvw'\) \(=u'vw+uv'w+uvw'\).

a) \(f'(x)=xe^x(2\ln(x)+x\ln(x)+1)\) b) Applying the product rule to \((uv)w\), then again to \(uv\), gives \((uvw)'=u'vw+uv'w+uvw'\).

- Write the product rule for three factors before substituting values. - Substitute each function value and derivative value carefully. - For four factors, how many terms should the derivative contain if exactly one factor is differentiated in each term?

1. For three factors, \(p'(x)=g'(x)h(x)k(x)+g(x)h'(x)k(x)+g(x)h(x)k'(x)\). 2. Substitute the values at \(x=5\): \(p'(5)=0.5 \cdot (-2) \cdot 10+4 \cdot 3 \cdot 10+4 \cdot (-2) \cdot (-1)\) \(=-10+120+8=118\). 3. For four factors, differentiate one factor at a time while leaving the other three unchanged: \((f_1f_2f_3f_4)'=f_1'f_2f_3f_4+f_1f_2'f_3f_4+f_1f_2f_3'f_4+f_1f_2f_3f_4'\).

a) \(p'(5)=118\) b) \((f_1f_2f_3f_4)'=f_1'f_2f_3f_4+f_1f_2'f_3f_4+f_1f_2f_3'f_4+f_1f_2f_3f_4'\)

- Identify the two factors in each function and differentiate them separately. - Use the product rule \(u'v+uv'\). - Watch the negative sign when differentiating cosine. - Rewrite \(\sqrt{x}\) as a power if helpful.

1. For a), let \(u(x)=x^2-4\) and \(v(x)=\cos(x)\). Then \(u'(x)=2x\) and \(v'(x)=-\sin(x)\). Therefore, \(f'(x)=2x\cos(x)-(x^2-4)\sin(x)\). 2. For b), let \(u(x)=\sqrt{x}\) and \(v(x)=e^x\). Then \(u'(x)=\frac{1}{2\sqrt{x}}\) and \(v'(x)=e^x\). Thus, \(g'(x)=\frac{e^x}{2\sqrt{x}}+\sqrt{x}\,e^x\). Using a common denominator, \(g'(x)=\frac{e^x(1+2x)}{2\sqrt{x}}\).

a) \(f'(x)=2x\cos(x)-(x^2-4)\sin(x)\) b) \(g'(x)=\frac{e^x(1+2x)}{2\sqrt{x}}\), for \(x>0\)

- Identify the two factors in each function. - Remember that the derivative of \(e^x\) is \(e^x\). - Look for a common factor after applying the product rule. - Check that each result follows the structure \(u'v+uv'\).

1. For a), let \(u(x)=x^3+2x\) and \(v(x)=\sin(x)\). Then \(u'(x)=3x^2+2\) and \(v'(x)=\cos(x)\). Therefore, \(h'(x)=(3x^2+2)\sin(x)+(x^3+2x)\cos(x)\). 2. For b), apply the product rule: \(k'(x)=e^x\cos(x)-e^x\sin(x)\). Factor out \(e^x\): \(k'(x)=e^x(\cos(x)-\sin(x))\).

a) \(h'(x)=(3x^2+2)\sin(x)+(x^3+2x)\cos(x)\) b) \(k'(x)=e^x(\cos(x)-\sin(x))\)

- Identify the two factors that depend on \(x\) in each function. - Treat parameters as constants when differentiating with respect to \(x\). - Recall the derivatives of sine and cosine. - Use parentheses carefully when combining terms and signs.

1. For a), treat \(k\) as a constant. The product rule gives \(f'(x)=2x\cos(x)-(x^2+k)\sin(x)\). 2. For b), treat \(a\) as a constant. Since \(\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\), \(g'(x)=\frac{a\sin(x)}{2\sqrt{x}}+a\sqrt{x}\cos(x)\). 3. For c), factor out the constant \(\frac{1}{t}\). Then \(h'(x)=\frac{1}{t}[\cos(x)-x\sin(x)]\) \(=\frac{\cos(x)-x\sin(x)}{t}\).

a) \(f'(x)=2x\cos(x)-(x^2+k)\sin(x)\) b) \(g'(x)=\frac{a\sin(x)}{2\sqrt{x}}+a\sqrt{x}\cos(x)\), for \(x>0\) c) \(h'(x)=\frac{\cos(x)-x\sin(x)}{t}\)

- Separate constants and parameters from factors that depend on \(x\). - Rewrite square roots as powers if helpful. - A constant in the denominator may be treated as a constant factor. - Simplify after applying the product rule.

1. For a), apply the product rule: \(f'(x)=-3x^2\sin(x)+(c-x^3)\cos(x)\). 2. For b), \(g'(x)=\frac{2x+c}{2\sqrt{x}}+2\sqrt{x}\). Using a common denominator, \(g'(x)=\frac{6x+c}{2\sqrt{x}}\). 3. For c), factor out \(\frac{1}{2c}\) and apply the product rule: \(h'(x)=\frac{\cos^2(x)-\sin^2(x)}{2c}\). Using \(\cos(2x)=\cos^2(x)-\sin^2(x)\), this may also be written as \(h'(x)=\frac{\cos(2x)}{2c}\).

a) \(f'(x)=-3x^2\sin(x)+(c-x^3)\cos(x)\) b) \(g'(x)=\frac{6x+c}{2\sqrt{x}}\), for \(x>0\) c) \(h'(x)=\frac{\cos^2(x)-\sin^2(x)}{2c}=\frac{\cos(2x)}{2c}\)

- How many factors make up the product? - Differentiate each factor before applying the three-factor product rule. - Differentiate one factor at a time in each term. - Factor out any common expression at the end.

1. Let \(u(x)=x^2+2\), \(v(x)=e^x\), and \(w(x)=\cos(x)\). 2. Then \(u'(x)=2x\), \(v'(x)=e^x\), and \(w'(x)=-\sin(x)\). 3. Apply the product rule for three factors: \(f'(x)=2xe^x\cos(x)+(x^2+2)e^x\cos(x)-(x^2+2)e^x\sin(x)\). 4. Factor out \(e^x\) and combine the cosine terms: \(f'(x)=e^x[(x^2+2x+2)\cos(x)-(x^2+2)\sin(x)]\).

\(f'(x)=e^x[(x^2+2x+2)\cos(x)-(x^2+2)\sin(x)]\)

- Differentiate each factor before combining the results. - Rewrite \(\sqrt{x}\) as a power if helpful. - Group terms containing the same trigonometric function. - A common denominator can help combine the sine terms.

1. Let \(u(x)=\sqrt{x}\), \(v(x)=\sin(x)\), and \(w(x)=x^2-4\). 2. Then \(u'(x)=\frac{1}{2\sqrt{x}}\), \(v'(x)=\cos(x)\), and \(w'(x)=2x\). 3. Apply the product rule for three factors: \(f'(x)=\frac{\sin(x)(x^2-4)}{2\sqrt{x}}+\sqrt{x}\cos(x)(x^2-4)+2x\sqrt{x}\sin(x)\). 4. Combine the sine terms using the common denominator \(2\sqrt{x}\): \(\frac{x^2-4+4x^2}{2\sqrt{x}}\sin(x)=\frac{5x^2-4}{2\sqrt{x}}\sin(x)\). Therefore, \(f'(x)=\frac{5x^2-4}{2\sqrt{x}}\sin(x)+(x^2-4)\sqrt{x}\cos(x)\).

\(f'(x)=\frac{5x^2-4}{2\sqrt{x}}\sin(x)+(x^2-4)\sqrt{x}\cos(x)\), for \(x>0\)

- Use the product rule for the two main factors. - Each factor is also a composite function. - Apply the chain rule carefully to \(\sin^3(2x)\). - Track the sign when differentiating cosine.

1. Let \(u(x)=\cos(x^2)\) and \(v(x)=\sin^3(2x)\). 2. By the chain rule, \(u'(x)=-2x\sin(x^2)\). Also, \(v'(x)=3\sin^2(2x)\cos(2x)\cdot 2=6\sin^2(2x)\cos(2x)\). 3. Apply the product rule: \(f'(x)=-2x\sin(x^2)\sin^3(2x)+6\cos(x^2)\sin^2(2x)\cos(2x)\).

\(f'(x)=-2x\sin(x^2)\sin^3(2x)+6\cos(x^2)\sin^2(2x)\cos(2x)\)

- Use the product rule for the two main factors. - Apply the chain rule to the cubic factor. - Look for a common power of \(3x+1\) before expanding. - Keep the final answer in factored form.

1. Let \(u(x)=x^2-4\) and \(v(x)=(3x+1)^3\). 2. Then \(u'(x)=2x\) and, by the chain rule, \(v'(x)=9(3x+1)^2\). 3. Apply the product rule: \(g'(x)=2x(3x+1)^3+9(x^2-4)(3x+1)^2\). 4. Factor out \((3x+1)^2\): \(g'(x)=(3x+1)^2[2x(3x+1)+9(x^2-4)]\). 5. Simplify the remaining factor: \(g'(x)=(3x+1)^2(15x^2+2x-36)\).

\(g'(x)=(3x+1)^2(15x^2+2x-36)\)

- Identify the two factors before applying the product rule. - Differentiate the exponential factor with the chain rule. - A local extremum occurs where the first derivative is zero. - A point on the y-axis has x-coordinate zero.

1. By the product rule, \(f_t'(x)=2e^{0.5x}+(2x-t)(0.5)e^{0.5x}=(x+2-0.5t)e^{0.5x}\). 2. Since \(e^{0.5x}>0\), the equation \(f_t'(x)=0\) reduces to \(x+2-0.5t=0\). Thus, the critical number is \(x=0.5t-2\). 3. The second derivative is \(f_t''(x)=(0.5x+2-0.25t)e^{0.5x}\). At \(x=0.5t-2\), it equals \(e^{0.25t-1}>0\), so the extremum is a local minimum. 4. A point on the y-axis has x-coordinate \(0\). Set \(0.5t-2=0\) to obtain \(t=4\).

1. \(f_t'(x)=(x+2-0.5t)e^{0.5x}\) 2. The x-coordinate is \(0.5t-2\). 3. \(t=4\)

- Identify the outer product and the composite functions inside it. - Use the chain rule for each factor. - Differentiate the two factors separately before applying the product rule. - Factor out the largest common power of \(x^2+1\).

1. Let \(u(x)=(x^2+1)^4\) and \(v(x)=\sin(5x)\). 2. By the chain rule, \(u'(x)=8x(x^2+1)^3\) and \(v'(x)=5\cos(5x)\). 3. Apply the product rule: \(f'(x)=8x(x^2+1)^3\sin(5x)+5(x^2+1)^4\cos(5x)\). 4. Factor out \((x^2+1)^3\): \(f'(x)=(x^2+1)^3[8x\sin(5x)+5(x^2+1)\cos(5x)]\).

\(f'(x)=(x^2+1)^3[8x\sin(5x)+5(x^2+1)\cos(5x)]\)

- Evaluate the function at \(x=0\). - Use the product rule and the chain rule. - The exponential factor is always positive. - Compare the x- and y-coordinates of the maximum.

1. For every \(t>0\), \(f_t(0)=0\), so all graphs pass through \((0, 0)\). If \(x\neq0\) were common to graphs with distinct parameters \(t\) and \(s\), then \(e^{1-x/t}=e^{1-x/s}\), which would imply \(t=s\). Thus the origin is the only common point. 2. By the product and chain rules, \(f_t'(x)=e^{1-x/t}\left(1-\frac{x}{t}\right)\). Since the exponential factor is positive, the only critical point is \(x=t\). 3. The derivative is positive for \(x<t\) and negative for \(x>t\), so this point is the unique local maximum. Its y-coordinate is \(f_t(t)=t\), giving the local maximum \((t, t)\). 4. Because \(x=t\) and \(y=t\), the locus is \(y=x\). The condition \(t>0\) gives \(x>0\).

a) \((0, 0)\) b) \((t, t)\) c) \(y=x\) for \(x>0\)

- Compare two functions with different parameter values to find common points. - Use the product rule and chain rule. - Factor the first derivative before solving. - Eliminate \(k\) from the maximum coordinates.

1. Every graph passes through \((0, 0)\). For \(x\neq0\), equality of two family members with distinct positive parameters would require \(e^{-k_1x}=e^{-k_2x}\), which is impossible unless \(k_1=k_2\). Thus the origin is the only common point. 2. By the product and chain rules, \(f_k'(x)=x(2-kx)e^{-kx}\). The critical points are \(x=0\) and \(x=\frac{2}{k}\). 3. The second derivative is \(f_k''(x)=(k^2x^2-4kx+2)e^{-kx}\). Since \(f_k''(0)=2>0\), the origin is a local minimum. At \(x=\frac{2}{k}\), \(f_k''\left(\frac{2}{k}\right)=-2e^{-2}<0\), so the local maximum is \(\left(\frac{2}{k}, \frac{4}{k^2e^2}\right)\). 4. At the maximum, \(x=\frac{2}{k}>0\), so \(k=\frac{2}{x}\). Substitution gives \(y=\frac{x^2}{e^2}\). Thus the locus is \(y=\frac{x^2}{e^2}\) for \(x>0\).

a) The origin is the only common point. b) Local minimum: \((0, 0)\); local maximum: \(\left(\frac{2}{k}, \frac{4}{k^2e^2}\right)\) c) \(y=\frac{x^2}{e^2}\) for \(x>0\)

- Compare exponential growth or decay with the linear factor. - Use the product rule for successive derivatives. - Eliminate \(a\) from each point’s coordinates.

1. As \(x\to\infty\), the factor \(a-x\) is negative and unbounded while \(e^x\) grows without bound, so \(g_a(x)\to-\infty\). As \(x\to-\infty\), exponential decay dominates the linear factor, so \(g_a(x)\to0\). 2. By the product rule, \(g_a'(x)=(a-1-x)e^x\) and \(g_a''(x)=(a-2-x)e^x\). The only critical point is \(x=a-1\). Since \(g_a''(a-1)=-e^{a-1}<0\), the local maximum is \((a-1, e^{a-1})\). 3. The inflection-point equation is \(g_a''(x)=0\), giving \(x=a-2\). The third derivative is nonzero there, so the inflection point is \((a-2, 2e^{a-2})\). 4. For the maxima, \(x=a-1\), so \(y=e^x\). For the inflection points, \(x=a-2\), so \(y=2e^x\).

a) \(\lim_{x\to\infty}g_a(x)=-\infty\) and \(\lim_{x\to-\infty}g_a(x)=0\) b) Local maximum: \((a-1, e^{a-1})\); inflection point: \((a-2, 2e^{a-2})\) c) Maximum locus: \(y=e^x\); inflection-point locus: \(y=2e^x\)

- The exponential factor never equals zero. - Use the product rule and the second derivative test. - Eliminate \(k\) from the minimum coordinates. - Set two family members equal to test for common points.

1. Since \(e^x\neq0\), the zero occurs at \(x=k\). By the product rule, \(f_k'(x)=e^x(x-k+1)\), so the critical point is \(x=k-1\). Also, \(f_k''(x)=e^x(x-k+2)\), and \(f_k''(k-1)=e^{k-1}>0\). Therefore, the local minimum is \((k-1, -e^{k-1})\). 2. At the minimum, \(x=k-1\), so \(k=x+1\). Substitution into the y-coordinate gives \(y=-e^x\). 3. If two distinct family members shared a point, then \((x-k)e^x=(x-m)e^x\). Since \(e^x\neq0\), this implies \(k=m\). Therefore, distinct graphs have no common points.

1. Zero: \(x=k\); local minimum: \((k-1, -e^{k-1})\) 2. \(y=-e^x\) 3. Distinct graphs have no common points.

- Use the domain condition when solving for a zero. - Apply the product rule and the derivative of \(\ln x\). - Express the maximum y-coordinate using its x-coordinate. - Compare two family members to test for common points.

1. Because \(x>0\), the equation \(f_k(x)=0\) reduces to \(k-\ln x=0\). Thus the zero is \(x=e^k\). 2. By the product rule, \(f_k'(x)=x(2k-1-2\ln x)\). Since \(x>0\), the critical-point equation gives \(\ln x=k-\frac{1}{2}\), so \(x=e^{k-1/2}\). 3. The second derivative is \(f_k''(x)=2k-3-2\ln x\). At the critical point it equals \(-2<0\), so the point is a local maximum. Its y-coordinate is \(\frac{1}{2}e^{2k-1}\), giving the local maximum \(\left(e^{k-1/2}, \frac{1}{2}e^{2k-1}\right)\). 4. Since \(x^2=e^{2k-1}\), the locus is \(y=\frac{1}{2}x^2\) for \(x>0\). 5. If two graphs with distinct parameters shared a point, then \(x^2(k_1-\ln x)=x^2(k_2-\ln x)\). This would require \(x=0\), which is outside the domain, or \(k_1=k_2\). Thus there are no common points.

a) \(x=e^k\) b) \(\left(e^{k-1/2}, \frac{1}{2}e^{2k-1}\right)\); locus: \(y=\frac{1}{2}x^2\) for \(x>0\) c) The graphs have no common points.

- Use the product rule and chain rule. - The exponential factor is always positive. - Use the second derivative test to classify the critical point. - Eliminate \(a\) from the maximum coordinates.

1. By the product rule, \(f_a'(x)=(1-a-x)e^{-x}\). Since \(e^{-x}>0\), the only critical point is \(x=1-a\). 2. The second derivative is \(f_a''(x)=(x+a-2)e^{-x}\). At the critical point, \(f_a''(1-a)=-e^{a-1}<0\), so this point is a local maximum. 3. Its y-coordinate is \(f_a(1-a)=e^{a-1}\), so the local maximum is \((1-a, e^{a-1})\). 4. Since \(x=1-a\), we have \(a=1-x\). Substitution gives \(y=e^{-x}\).

a) \((1-a, e^{a-1})\) b) \(y=e^{-x}\)

- Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - A line through the origin has the form \(y=mx\). - At a tangency point, the secant slope from the origin equals the derivative. - Intersection points satisfy equal function values.

1. Since \(f(-x)=(-x)\cos(-x)=-x\cos(x)=-f(x)\), the function is odd and the graph has origin symmetry. The zeros satisfy \(x\cos(x)=0\), giving \(x=-\frac{\pi}{2},0,\frac{\pi}{2}\). The y-intercept is also \((0,0)\). 2. A line through the origin has equation \(h(x)=mx\). At a tangency point with \(x_0\ne0\), its slope must satisfy \(m=\frac{f(x_0)}{x_0}=\cos(x_0)\) and \(m=f'(x_0)=\cos(x_0)-x_0\sin(x_0)\). Therefore, \(x_0\sin(x_0)=0\). With \(x_0>0\) in the stated interval, \(x_0=\pi\). Then \(P=(\pi,-\pi)\) and \(m=-1\), so \(h(x)=-x\). 3. Solve \(x\cos(x)=\frac{1}{2}x\): \(x\left(\cos(x)-\frac{1}{2}\right)=0\). Thus \(x=0\) or \(\cos(x)=\frac{1}{2}\). In \([-\pi,\pi]\), the solutions are \(x=0,\pm\frac{\pi}{3}\). The intersection points are \((0,0)\), \(\left(\frac{\pi}{3},\frac{\pi}{6}\right)\), and \(\left(-\frac{\pi}{3},-\frac{\pi}{6}\right)\).

a) The graph has origin symmetry. Its intercepts are \(\left(-\frac{\pi}{2},0\right)\), \((0,0)\), and \(\left(\frac{\pi}{2},0\right)\). b) \(P=(\pi,-\pi)\), and \(h(x)=-x\). c) \((0,0)\), \(\left(\frac{\pi}{3},\frac{\pi}{6}\right)\), and \(\left(-\frac{\pi}{3},-\frac{\pi}{6}\right)\).

- Apply the product rule and chain rule. - Use the nonzero critical point for the local maximum. - Eliminate \(k\) from the maximum coordinates. - Compare the computed maximum x-coordinates with the graph.

1. By the product and chain rules, \(f_k'(x)=x(2-kx)e^{-kx}\). For \(x>0\), the local maximum occurs at \(x=\frac{2}{k}\). 2. If \(x=1\), then \(k=2\). The y-coordinate is \(f_2(1)=e^{-2}\), so the maximum is \((1, e^{-2})\). 3. At a maximum, \(k=\frac{2}{x}\). Substitution gives \(y=x^2e^{-2}\), with \(x>0\). 4. For \(k=0.5\), the maximum occurs at \(x=4\), matching p. For \(k=1.0\), it occurs at \(x=2\), matching q.

a) \(k=2\); maximum \((1, e^{-2})\) b) \(y=e^{-2}x^2\) for \(x>0\) c) p: \(k=0.5\); q: \(k=1.0\)

- Use the y-intercept to find one parameter first. - Apply the product rule and chain rule. - The derivative is zero at a local extremum.

1. The y-intercept condition gives \(f(0)=b=2\). 2. By the product rule, \(f'(x)=(a-0.5ax-0.5b)e^{-0.5x}\). 3. A local maximum at \(x=1\) requires \(f'(1)=0\). Substituting \(b=2\) gives \((0.5a-1)e^{-0.5}=0\), so \(a=2\). 4. The derivative changes from positive to negative at \(x=1\), confirming the local maximum.

\(a=2\), \(b=2\)

- Substitute the point into the function. - Apply the product rule before evaluating the derivative. - Remember both square-root solutions.

1. Substituting \(x=-1\) gives \(f_k(-1)=(1-k^2)e^{-1}\). This equals zero when \(k^2=1\), so \(k=\pm1\). 2. The product rule gives \(f_k'(x)=(k^2x+k^2+1)e^x\). 3. Thus \(f_k'(0)=k^2+1\). Setting this equal to \(5\) gives \(k^2=4\), so \(k=\pm2\).

1. \(k=\pm1\) 2. \(k=\pm2\)

- A horizontal tangent has slope zero. - Use the product rule and chain rule. - The exponential factor cannot be zero.

1. By the product and chain rules, \(f_k'(x)=e^{-0.5x}(1-0.5x+0.5k)\). 2. A horizontal tangent at \(x=2\) requires \(f_k'(2)=0\). 3. This gives \(e^{-1}(0.5k)=0\). Since \(e^{-1}\ne0\), \(k=0\).

\(k=0\)

- The exponential factor is never zero. - Use the product rule and chain rule. - Compare the two x-coordinate expressions. - An exponential equals \(1\) when its exponent is zero.

1. Since the exponential factor is positive, the zero is \(x=k\). 2. The derivative is \(f_k'(x)=(1-0.5x+0.5k)e^{1-0.5x}\). Setting it equal to zero gives \(x=k+2\). The derivative changes from positive to negative, so this is a local maximum. 3. The distance between the zero and the local maximum is \((k+2)-k=2\). 4. The slope at the zero is \(f_k'(k)=e^{1-0.5k}\). Setting this equal to \(1\) gives \(1-0.5k=0\), so \(k=2\).

a) \(x=k\) b) Local maximum x-coordinate: \(x=k+2\); the distance from the zero is \(2\). c) \(k=2\)

- Treat \(f_n(x) = f_{n-1}(x)(x-(n+1))\) as a product of two functions. - Notice how one additional factor creates one additional term in the derivative. - In the general formula, each product should contain all but one of the original factors.

1. Expanding \(f_1\) gives \(f_1(x) = x^2 - 3x + 2\), so \(f_1'(x) = 2x - 3\). For \(f_2\), differentiate one factor at a time: \(f_2'(x) = (x-2)(x-3) + (x-1)(x-3) + (x-1)(x-2)\) \(= 3x^2 - 12x + 11\). 2. Write \(f_n(x) = f_{n-1}(x)v(x)\), where \(v(x) = x-(n+1)\) and \(v'(x)=1\). The product rule gives \(f_n'(x) = f_{n-1}'(x)v(x) + f_{n-1}(x)v'(x)\) \(= f_{n-1}'(x)(x-(n+1)) + f_{n-1}(x)\). 3. Repeating the product rule shows that each term in \(g'(x)\) omits exactly one original factor: \(g'(x) = \sum_{i=1}^{k}\left(\prod_{\substack{1 \le j \le k \\ j \ne i}}(x-a_j)\right)\).

1. \(f_1'(x) = 2x - 3\); \(f_2'(x) = 3x^2 - 12x + 11\) 2. \(f_n'(x) = f_{n-1}'(x)(x-(n+1)) + f_{n-1}(x)\) 3. \(g'(x) = \sum_{i=1}^{k}\left(\prod_{\substack{1 \le j \le k \\ j \ne i}}(x-a_j)\right)\)

- Total equals quantity times average. - Apply the product rule to \(xk(x)\). - Use the sign of \(k'(x)\) and the fact that \(x>0\). - Think about whether a new value must be above or below the current average to change that average downward.

1. Apply the product rule to \(K(x)=xk(x)\): \(K'(x)=k(x)+xk'(x)\). 2. If \(k'(x)<0\) and \(x>0\), then \(xk'(x)<0\). 3. Therefore, \(K'(x)=k(x)+xk'(x)<k(x)\). 4. For test scores, a new score below the current average pulls the average downward; a new score above the current average pushes it upward. Marginal cost plays the role of the next value being added.

a) \(K'(x)=k(x)+xk'(x)\) b) Since \(x>0\) and \(k'(x)<0\), \(xk'(x)<0\), so \(K'(x)<k(x)\). c) A new test score below the current average lowers the average score.

- Use the exponential factor's sign when finding zeros and critical points. - The area in the fourth quadrant is the negative of the definite integral. - Write the tangent line at a general value \(u\) and substitute \(P(k, 0)\). - A quadratic has two distinct real solutions when its discriminant is positive.

1. Since \(e^x>0\), \(f(x)=0\) when \(x-2=0\), so the zero is \(x=2\). 2. Differentiate using the product rule: \(f'(x)=(x-1)e^x\). The only critical point is \(x=1\). Since \(f''(1)=e>0\), the graph has a local minimum at \(T(1, -e)\). 3. On \([0, 2]\), \(f(x)\leq0\). An antiderivative is \((x-3)e^x\), so the enclosed area is \(A=-\int_0^2(x-2)e^x\,dx=-\left[(x-3)e^x\right]_0^2=e^2-3\approx4.39\). 4. The tangent line at \(x=u\) is \(y=(u-1)e^u(x-u)+(u-2)e^u\). Requiring it to pass through \(P(k, 0)\) gives \(0=(u-1)e^u(k-u)+(u-2)e^u\). 5. Divide by \(e^u>0\) and simplify: \(0=(u-1)(k-u)+u-2=-u^2+(k+2)u-(k+2)\). Therefore, \(u^2-(k+2)u+(k+2)=0\). 6. This quadratic has two distinct real solutions when its discriminant is positive: \(\Delta=(k+2)^2-4(k+2)=k^2-4>0\). Thus \(k<-2\) or \(k>2\). The two solutions give distinct points of tangency and distinct tangent lines.

a) Zero: \(x=2\); local minimum: \(T(1, -e)\) b) \(A=e^2-3\approx4.39\) square units c) \(u^2-(k+2)u+(k+2)=0\) d) \(k<-2\) or \(k>2\)