Free math worksheets

- Differentiate both sides with respect to \(x\). - Treat \(y\) as a function of \(x\), not as a constant. - The derivative of the constant on the right side is \(0\).

1. Differentiate both sides with respect to \(x\): \(1+\frac{dy}{dx}=0\). 2. Therefore, \(\frac{dy}{dx}=-1\).

\(\frac{dy}{dx}=-1\)

- Substitute the point into the differentiated equation first. - After substitution, \(\frac{dy}{dx}\) is the only unknown. - Solve the resulting linear equation for the slope.

1. Substitute \(x=3\) and \(y=4\): \(6+8\frac{dy}{dx}=0\). 2. Solving gives \(\frac{dy}{dx}=-\frac{3}{4}\).

\(-\frac{3}{4}\)

- Treat \(y\) as a function whose input is \(x\), not as a constant. - For a power of \(y\), think of the power as the outer function and \(y(x)\) as the inner function. - When \(x\) and \(y\) are multiplied, both factors depend on the differentiation process.

1. a) Because \(y=y(x)\), use the chain rule: \(\frac{d}{dx}(y^2)=2y y^{\prime}\). The factor \(y^{\prime}\) comes from differentiating the inner function \(y(x)\). 2. b) Differentiate term by term: \(\frac{d}{dx}(3x+y^3)=3+3y^2y^{\prime}\). The factor \(y^{\prime}\) comes from the chain rule on \(y^3\). 3. c) Use the product rule: \(\frac{d}{dx}(xy)=y+xy^{\prime}\). The factor \(y^{\prime}\) appears when differentiating the factor \(y(x)\).

a) \(2y y^{\prime}\); chain rule b) \(3+3y^2y^{\prime}\); chain rule on \(y^3\) c) \(y+xy^{\prime}\); product rule, with \(y^{\prime}\) from differentiating \(y(x)\)

- Differentiate both sides of the equation with respect to \(x\). - When differentiating a power of \(y\), remember that \(y\) depends on \(x\). - Isolate \(\frac{dy}{dx}\) before substituting the coordinates of the point.

1. a) Differentiate both sides with respect to \(x\): \(2x+4y\frac{dy}{dx}=0\). Solve for the derivative: \(\frac{dy}{dx}=-\frac{x}{2y}\). 2. b) The point lies on the curve because \(4^2+2(1)^2=18\). Substitute \(x=4\) and \(y=1\): \(\frac{dy}{dx}=-\frac{4}{2(1)}=-2\).

a) \(\frac{dy}{dx}=-\frac{x}{2y}\) b) \(-2\)

- Treat the term \(xy\) as a product of two factors before differentiating. - Keep every term containing \(\frac{dy}{dx}\) together after differentiating. - Substitute the point only after you have a derivative formula in terms of \(x\) and \(y\).

1. a) Differentiate both sides with respect to \(x\). The product rule gives \(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\), and the chain rule gives \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\). Thus, \(y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\). Collect the derivative terms and solve: \(\frac{dy}{dx}=-\frac{y}{x+2y}\). 2. b) At \((1,2)\), \(\frac{dy}{dx}=-\frac{2}{1+4}=-\frac{2}{5}\). 3. c) Both \(x\) and \(y=y(x)\) vary with \(x\), so the product rule differentiates each factor in turn. The missing term would be \(x\frac{dy}{dx}\).

a) \(\frac{dy}{dx}=-\frac{y}{x+2y}\) b) \(-\frac{2}{5}\) c) Both factors in \(xy\) vary with \(x\), so the product rule gives \(y+x\frac{dy}{dx}\), not just \(y\).

- Differentiate the equation before substituting the coordinates of the point. - A power of \(y\) requires a chain-rule factor involving \(\frac{dy}{dx}\). - Once you have the slope, use the given point to write a tangent-line equation.

1. Differentiate implicitly: \(2x+3y^2\frac{dy}{dx}=0\). 2. Solve for the derivative: \(\frac{dy}{dx}=-\frac{2x}{3y^2}\). 3. At \((1,2)\), the slope is \(-\frac{2}{3(4)}=-\frac{1}{6}\). 4. Use point-slope form: \(y-2=-\frac{1}{6}(x-1)\).

\(y-2=-\frac{1}{6}(x-1)\)

- Ask whether \(y\) is being treated as a constant or as a function of \(x\). - When a trigonometric function has \(y\) as its input, differentiate the outer trig function and then the inner function. - Check that the given point satisfies the original equation before evaluating the derivative.

1. a) The student treated \(y\) as if it were independent of \(x\). Since \(y=y(x)\), differentiating \(\sin y\) requires the chain rule and a factor of \(\frac{dy}{dx}\). 2. b) The correct differentiated equation is \(2x+\cos y\frac{dy}{dx}=0\). Therefore, \(\frac{dy}{dx}=-\frac{2x}{\cos y}\), wherever \(\cos y\ne0\). 3. c) The point lies on the curve because \(1^2+\sin 0=1\). At \((1,0)\), \(\frac{dy}{dx}=-\frac{2}{1}=-2\).

a) The derivative of \(\sin y\) is missing the chain-rule factor \(\frac{dy}{dx}\). b) \(\frac{dy}{dx}=-\frac{2x}{\cos y}\) c) \(-2\)

- Use the grid and marker to determine the point before differentiating. - Differentiate both squared terms with respect to \(x\), remembering that \(y=y(x)\). - Use the derivative value as the tangent slope in point-slope form.

1. a) From the grid, \(P=(2,2)\). 2. b) Differentiate the curve equation: \(2x+8y\frac{dy}{dx}=0\). Thus \(\frac{dy}{dx}=-\frac{x}{4y}\). At \((2,2)\), the slope is \(-\frac{2}{8}=-\frac{1}{4}\). 3. c) Using point-slope form, the tangent line is \(y-2=-\frac{1}{4}(x-2)\).

a) \((2,2)\) b) \(-\frac{1}{4}\) c) \(y-2=-\frac{1}{4}(x-2)\)

- First solve the differentiated equation for \(\frac{dy}{dx}\). - A horizontal tangent is controlled by the numerator of the derivative, provided the denominator is nonzero. - A vertical tangent is controlled by the denominator, provided the numerator is nonzero. - Check the original curve equation after imposing each tangent condition.

1. Differentiate implicitly: \(2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\). Thus \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\). 2. a) A horizontal tangent requires \(2x+y=0\) and \(x+2y\ne0\). Set \(y=-2x\). Substitution into the curve gives \(3x^2=7\), so \(x=\pm\sqrt{\frac{7}{3}}\). The points are \(\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right)\) and \(\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)\). At these points, \(x+2y=-3x\ne0\). 3. b) A vertical tangent requires \(x+2y=0\) and \(2x+y\ne0\). Set \(x=-2y\). Substitution gives \(3y^2=7\), so \(y=\pm\sqrt{\frac{7}{3}}\). The points are \(\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right)\) and \(\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)\). At these points, \(2x+y=-3y\ne0\).

a) \(\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right)\) and \(\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)\) b) \(\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right)\) and \(\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)\)

- Differentiate the original equation once before thinking about the second derivative. - When differentiating a product involving \(y\) and \(\frac{dy}{dx}\), both factors depend on \(x\). - Keep \(\frac{d^2y}{dx^2}\) symbolic until you substitute the point. - Use the first derivative value at the point in the second differentiated equation.

1. Differentiate once: \(2x+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x}{y}\). 2. Differentiate the equation \(2x+2y\frac{dy}{dx}=0\) again: \(2+2\left(\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right)=0\). 3. At \((3,4)\), \(\frac{dy}{dx}=-\frac{3}{4}\). Substitute this value: \(2+2\left(\frac{9}{16}+4\frac{d^2y}{dx^2}\right)=0\). 4. Solving gives \(\frac{d^2y}{dx^2}=-\frac{25}{64}\).

\(\frac{d^2y}{dx^2}=-\frac{25}{64}\)

- The parameter is constant when differentiating with respect to \(x\). - Find the implicit derivative before substituting the fixed point. - Parallel lines have equal slopes, so translate the line condition into an equation involving the parameter.

1. Differentiate implicitly with respect to \(x\): \(2x+2y\frac{dy}{dx}+a=0\). 2. Solve for the slope: \(\frac{dy}{dx}=-\frac{2x+a}{2y}\). 3. At \((1,2)\), the slope is \(-\frac{2+a}{4}\). 4. A line parallel to \(y=-x+6\) has slope \(-1\). Set \(-\frac{2+a}{4}=-1\). Then \(2+a=4\), so \(a=2\).

\(a=2\)

- Differentiate every occurrence of \(y\) as a function of \(x\). - After differentiating, collect all terms containing \(\frac{dy}{dx}\) on one side. - For the method comparison, focus on what information is needed to find a slope at one point rather than a formula for the entire curve.

1. a) Differentiate both sides with respect to \(x\): \(3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}\). Collect the derivative terms: \(\left(3y^2-6x\right)\frac{dy}{dx}=6y-3x^2\). Thus \(\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}\). At \((3,3)\), \(\frac{dy}{dx}=\frac{18-27}{27-18}=-1\). 2. b) Implicit differentiation works directly with the relation that the point satisfies and does not require choosing or solving for a global explicit branch. At \((3,3)\), the coefficient \(3y^2-6x=9\ne0\), so the differentiated equation determines a unique finite slope there.

a) \(-1\) b) Implicit differentiation uses the given relation directly and avoids solving a cubic for \(y\); at \((3,3)\), the derivative equation has a nonzero coefficient on \(\frac{dy}{dx}\), so it determines the local slope.