- A product derivative contains two terms. - In each term, exactly one factor is differentiated. - The product of the two derivatives alone is not the product rule.

1. The derivative of a product is the derivative of the first factor times the unchanged second factor plus the unchanged first factor times the derivative of the second factor. 2. Therefore, formula B is the product rule.

B) \(u'(x)v(x)+u(x)v'(x)\)

- Write the product rule before substituting any numerical values. - Each product-rule term uses one function value and the other function's derivative value. - Keep the negative function value attached to its factor.

1. By the product rule, \(h'(2)=u'(2)v(2)+u(2)v'(2)\). 2. Substitute the given values: \(h'(2)=4(-1)+3(5)=11\).

\(h'(2)=11\)

- Ask which parts of each expression change when \(x\) changes. - Can the sum rule alone differentiate both expressions directly?

1. In \(f\), the two variable-dependent factors \(x\) and \(e^x\) are multiplied, so the product rule applies. 2. In \(g\), the two terms are added, so the sum rule applies.

\(f(x)=xe^x\), because it multiplies two functions of \(x\).

- Treat \(x\) and \(x+1\) as separate factors. - The product rule gives two added terms, not the product of the derivatives.

1. Differentiate the first factor and leave the second, then switch roles: \(f'(x)=1(x+1)+x(1)\). 2. Simplify: \(f'(x)=2x+1\).

\(f'(x)=1(x+1)+x(1)=2x+1\)

- 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 \((uv)'(x)\). - Use the given derivative of \(v\) and differentiate \(u\). - Compute both proposed expressions and simplify before comparing them. - Does agreement for one pair of functions prove a rule for every pair?

1. We have \(u'(x)=1\), and the derivative of \(v\) is supplied. 2. By 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 proposed expression gives \(u'(x)v'(x)=\frac{1}{(1-x)^2}\). They are equal for every \(x\ne1\). The claim is true for this pair, but not as a general rule.

The claim is true for these functions because both sides equal \(\frac{1}{(1-x)^2}\) for every \(x\ne1\).

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

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

- Identify the two factors in each function before differentiating. - In a product derivative, each original factor appears once without being differentiated. - Use the established derivatives of \(\sin x\) and \(\ln x\) for the individual factors.

1. For part a, apply the product rule: \(f'(x)=1\cdot\sin x+x\cos x=\sin x+x\cos x\). 2. For part b, apply the product rule: \(g'(x)=2\ln x+(2x-5)\frac{1}{x}\). 3. Simplifying gives \(g'(x)=2\ln x+2-\frac{5}{x}\).

a) \(f'(x)=\sin x+x\cos x\) b) \(g'(x)=2\ln x+2-\frac{5}{x}\)

- 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 \(x>0\)

- What is the derivative of a constant? - Substitute the two factors into the product rule. - What happens to the first product-rule term when one factor is constant? - Which method requires fewer steps?

1. By the constant multiple rule, \(f'(x)=8\cos(x)\). 2. Using the product rule, \(u'(x)=0\) and \(v'(x)=\cos(x)\), so \(f'(x)=u'(x)v(x)+u(x)v'(x)\) \(=0\sin(x)+8\cos(x)=8\cos(x)\). 3. The methods agree because the derivative of the constant factor is zero. The constant multiple rule is more efficient because it omits the product-rule term that equals zero.

1. \(f'(x)=8\cos(x)\) 2. \(f'(x)=8\cos(x)\). The product rule gives the same result because the derivative of \(8\) is \(0\); the constant multiple rule is more efficient because it avoids writing the zero product-rule term.

- Use the product rule. - Recall the derivative of \(3^x\). - Evaluate the derivative at \(x=0\). - Use the origin and the slope to write the tangent line.

1. Use the product rule with \(u(x)=x\) and \(v(x)=3^x\): \(g'(x)=3^x+x(\ln 3)3^x\) \(=3^x(1+x\ln 3)\). 2. Since \(g(0)=0\), the graph passes through the origin. The tangent slope is \(g'(0)=3^0(1+0\ln 3)=1\). Therefore, the tangent line is \(y=x\).

a) \(g'(x)=3^x(1+x\ln 3)\) b) \(y=x\)

- Use the point condition independently of the derivative condition. - Differentiate the exponential factor with its inner function in mind. - Evaluate the derivative only after simplifying the product-rule result.

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

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

- Match each entry in the table to the corresponding factor in the Product Rule. - The derivative of a product uses both function values and both derivative values. - Substitute only after writing the Product Rule in symbols.

1. Use the Product Rule: \(h'(x)=f'(x)g(x)+f(x)g'(x)\). 2. Substitute the table values at \(x=2\): \(h'(2)=5(-4)+3(2)=-14\).

\(h'(2)=-14\)

- Think about how a derivative of a product differs from a product of derivatives. - Which given function value may make part of your calculation simpler?

1. The product rule gives \(h'(1)=f'(1)g(1)+f(1)g'(1)\). 2. Since \(f(1)=0\), the second term vanishes, leaving \(h'(1)=3\cdot7=21\).

\(h'(1)=f'(1)g(1)+f(1)g'(1)=3\cdot7+0\cdot g'(1)=21\). The unknown derivative is multiplied by zero.

- The area uses two dimensions that both depend on time. - Differentiate their product before substituting the time. - Area changes per unit time, so the answer's units are not just centimeters.

1. The area is \(A(t)=L(t)W(t)=(t+2)(t+1)\). 2. By the product rule, \(A'(t)=1(t+1)+(t+2)1=2t+3\). 3. At \(t=1\), \(A'(1)=5\,\text{cm}^2/\text{min}\).

\(A'(t)=L'(t)W(t)+L(t)W'(t)=2t+3\); at \(t=1\), \(5\,\text{cm}^2/\text{min}\).

- The product derivative combines two terms, only one of which is unknown here. - Substitute known values into the full rule before solving.

1. By the product rule, \(17=f'(2)g(2)+f(2)g'(2)=5+2g'(2)\). 2. Solve to obtain \(2g'(2)=12\), so \(g'(2)=6\).

\(g'(2)=6\)

- 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 \((-4, 8e^{-4})\)

- 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. Apply the product rule to \(k\): \(k'(x)=2xf(x)+x^2f'(x)\). Then \(k'(1)=2\cdot1\cdot(-2)+1^2\cdot0=-4\), so \(k\) does not have a horizontal tangent at \(x=1\). 3. Apply the product rule to \(m\): \(m'(x)=f'(x)e^x+(f(x)+2)e^x\). Then \(m'(1)=0\cdot e+(-2+2)e=0\), so \(m\) has 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.

- Evaluate the function at the given input. - Use the product rule to find the derivative. - Simplify with a double-angle identity if helpful. - Use negative reciprocal slopes for tangent and normal lines.

1. Evaluate the function: \(f\left(\frac{\pi}{2}\right)=1\). Thus, the point is \(\left(\frac{\pi}{2}, 1\right)\). 2. Differentiate using the product rule: \(f'(x)=\cos x(1+\cos x)-\sin^2x\). Using \(\cos^2x-\sin^2x=\cos(2x)\), \(f'(x)=\cos x+\cos(2x)\). Therefore, \(f'\left(\frac{\pi}{2}\right)=-1\). 3. The tangent line is \(y-1=-\left(x-\frac{\pi}{2}\right)\), or \(y=-x+\frac{\pi}{2}+1\). 4. The normal slope is \(1\), so the normal line is \(y-1=x-\frac{\pi}{2}\), or \(y=x-\frac{\pi}{2}+1\).

Tangent line: \(y=-x+\frac{\pi}{2}+1\) Normal line: \(y=x-\frac{\pi}{2}+1\)

- 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) \(k=4\)

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

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

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

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

- Check the zero-of-the-function statement directly from the product definition. - Derive \(f'(x)\) before deciding the derivative statements. - Distinguish carefully between knowing \(u(x_0)=0\) and knowing \(u'(x_0)=0\).

1. Statement (1) is true because \(f(x_0)=u(x_0)\ln x_0=0\). 2. Apply the Product Rule: \(f'(x)=u'(x)\ln x+\frac{u(x)}{x}\). If \(u'(x_0)=0\), then \(f'(x_0)=\frac{u(x_0)}{x_0}\), which need not be zero. Statement (2) is false. 3. If both \(u(x_0)=0\) and \(u'(x_0)=0\), substituting those values into the Product Rule formula gives \(f'(x_0)=0\). Statement (3) is true.

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

- Determine each factor's domain before differentiating. - Apply the product rule to both functions. - Use the established derivatives of \(e^x\) and \(\ln x\) for the individual factors. - Simplify only after the product-rule terms are complete.

1. In part a), both factors are defined for every real \(x\), so the domain is \(\mathbb{R}\). By the product rule, \(f'(x)=2xe^x+(x^2+1)e^x=(x+1)^2e^x\). 2. In part b), the logarithm requires \(x>0\), so the domain is \((0,\infty)\). By the product rule, \(g'(x)=2x\ln x+(x^2+1)\frac{1}{x}\). 3. Simplify: \(g'(x)=2x\ln x+x+\frac{1}{x}\).

a) Domain: \(\mathbb{R}\); \(f'(x)=(x+1)^2e^x\) b) Domain: \((0,\infty)\); \(g'(x)=2x\ln x+x+\frac{1}{x}\)

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

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

- The sine term is a product of a polynomial and a composite trigonometric function. - Identify the input of the sine before differentiating its factor. - Keep track of the subtraction sign in the final combination.

1. Use the product rule on \(3x^2\sin(x^2+1)\). The factor derivatives are \(6x\) and \(2x\cos(x^2+1)\). 2. Thus the first term's derivative is \(6x\sin(x^2+1)+6x^3\cos(x^2+1)\). 3. By the chain rule, the derivative of \(\frac14(2x-5)^8\) is \(4(2x-5)^7\). 4. Apply the subtraction sign from the original expression.

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

- 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.

- First find the input that makes the parameter-dependent part vanish. - Expand the shifted square before differentiating its polynomial factor. - Use the product rule only after the derivative of that factor is available.

1. The parameter term is zero when \(x=1\), so \(g_t(1)=e\). Every graph passes through \(Q(1,e)\). 2. Expand the square inside the polynomial factor: \(t(x-1)^2+1=t(x^2-2x+1)+1\). 3. Differentiate the polynomial factor with the power and sum rules: its derivative is \(t(2x-2)\). 4. Apply the product rule: \(g_t'(x)=e^x\left[t(2x-2)+t(x^2-2x+1)+1\right]=e^x\left[t(x^2-1)+1\right]\). 5. At \(x=1\), \(g_t'(1)=e\), independent of \(t\). 6. 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. a) \(k'(x)=6x(x^2+1)^2\) b) \(k'(x)=6x^5+12x^3+6x\). The two results are equivalent.

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

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

- What derivative value corresponds to a horizontal tangent? - Use both the product rule and the chain rule. - Include the derivative of the inner expression \(c-x\). - Substitute \(x=\frac{c}{2}\) before doing unnecessary simplification.

1. Differentiate using the product and chain rules: \(f'(x)=\sec^2(x)\tan(c-x)-\tan(x)\sec^2(c-x)\). 2. Substitute \(x=\frac{c}{2}\): \(f'\left(\frac{c}{2}\right)=\sec^2\left(\frac{c}{2}\right)\tan\left(\frac{c}{2}\right)-\tan\left(\frac{c}{2}\right)\sec^2\left(\frac{c}{2}\right)=0\). 3. Therefore, the tangent slope is \(0\), so the graph has a horizontal tangent at \(x=\frac{c}{2}\). The condition that \(f\) is defined there is equivalent to \(\cos\left(\frac{c}{2}\right)\neq0\).

\(f'\left(\frac{c}{2}\right)=0\), so the tangent is horizontal whenever \(\cos\left(\frac{c}{2}\right)\neq0\).

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

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

- Find the point on the graph at \(x=1\). - Apply the product rule. - Evaluate the derivative at \(x=1\). - Use point-slope form.

1. Evaluate the function: \(f(1)=1^2\ln1=0\). 2. Differentiate using the product rule: \(f'(x)=2x\ln x+x\). At \(x=1\), \(f'(1)=1\). 3. The tangent line through \((1, 0)\) with slope \(1\) is \(y=x-1\).

\(y=x-1\)

- 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 \(\left(-\frac{\pi}{3},-\frac{\pi}{6}\right)\).

- Substitute the given point condition directly into the family. - Apply the product rule before evaluating the derivative condition. - A squared parameter equation can have two real solutions.

1. Substituting \(x=-1\) gives \(f_k(-1)=(1-k^2)e^{-1}\). This is zero when \(k^2=1\), so \(k=\pm1\). 2. By the product rule, \(f_k'(x)=(k^2x+k^2+1)e^x\). 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 gives a zero-derivative condition. - Use both the product rule and the chain rule when differentiating. - The exponential factor itself cannot make the derivative 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 \(e^{-1}\ne0\), \(k=0\).

\(k=0\)

- 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. \(g'(x) = \sum_{i=1}^{k}\left(\prod_{\substack{1 \le j \le k \\ j \ne i}}(x-a_j)\right)\)

- Use the y-intercept condition before differentiating. - Relate the stated angle with the y-axis to a line's angle of inclination. - Evaluate the derivative at the y-intercept and compare that slope with the possible line directions. - Remember to enforce the restriction on \(k\).

1. The y-intercept gives \(f_{a,k}(0)=a=3\). 2. Differentiate using the product and chain rules: \(f_{a,k}'(x)=(1+kx+ak)e^{kx}\). 3. At \(x=0\), the tangent slope is \(1+3k\). 4. A line making a \(45^\circ\) angle with the y-axis also makes a \(45^\circ\) or \(135^\circ\) angle of inclination with the x-axis, so its slope is \(1\) or \(-1\). 5. The equation \(1+3k=1\) gives \(k=0\), which is excluded. The equation \(1+3k=-1\) gives \(k=-\frac{2}{3}\). 6. Therefore, \(a=3\) and \(k=-\frac{2}{3}\).

\(a=3\) and \(k=-\frac{2}{3}\)

- 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.

- Write the tangent line at a general input using the function value and derivative there. - Substitute the fixed x-axis point into that tangent equation. - After simplifying to a quadratic in the tangency input, use the condition for two distinct real roots.

1. By the product rule, \(f'(x)=(x-1)e^x\). 2. The tangent line at \(x=u\) is \(y=(u-1)e^u(x-u)+(u-2)e^u\). 3. Requiring the tangent to pass through \(P(k,0)\) gives \(0=(u-1)e^u(k-u)+(u-2)e^u\). 4. Since \(e^u>0\), divide by \(e^u\) and simplify to obtain \(u^2-(k+2)u+(k+2)=0\). 5. There are exactly two distinct real values of \(u\) when the discriminant is positive: \((k+2)^2-4(k+2)=k^2-4>0\). 6. Therefore, \(k<-2\) or \(k>2\).

a) \(u^2-(k+2)u+(k+2)=0\) b) \(k<-2\) or \(k>2\)