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AP Calculus BC : Chain Rule and Implicit Differentiation

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Example Questions

Example Question #51 : Computation Of Derivatives

Find :

$f=\gamma \cos(x^2)$ , where $\gamma$ is a constant.

Possible Answers:

$2x\gamma \cos(x^2)$

$-\gamma \sin(x^2)$

$2x\gamma \sin(x^2)$

$-2x\gamma \sin(x^2)$

$-2x\sin(x^2)$

Correct answer:

$-2x\gamma \sin(x^2)$

Explanation:

The derivative of the function is equal to

$f'=-2x\gamma \sin(x^2)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The constant may seem intimidating, but we treat it as another constant!

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Example Question #61 : Computation Of Derivatives

Find $\frac{\mathrm{d} \delta}{\mathrm{d} x}$ from the following equation:

$5=\delta e^{x^2}$ , where $\delta$ is a function of x.

Possible Answers:

-2 x

$-2\delta x$

$-2\delta x e^{x^2}$

$2\delta x$

Correct answer:

$-2\delta x$

Explanation:

To find the derivative of $\delta$ with respect to x, we must differentiate both sides of the equation with respect to x:

$0=2\delta xe^{x^2}+e^{x^2}\frac{\mathrm{d} \delta}{\mathrm{d} x}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Solving for $\frac{\mathrm{d} \delta}{\mathrm{d} x}$ , we get

$\frac{\mathrm{d} \delta}{\mathrm{d} x}=\frac{-2\delta x e^{x^2}}{e^{x^2}}=-2\delta x$

Note that the chain rule was used because of the exponential and because $\delta$ is a function of x.

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Example Question #161 : Derivatives

Find the first derivative of the following function:

$f=\sec(x^2)+\ln(2x^2)$

Possible Answers:

$\sec(x^2)\tan(x^2)+\frac{1}{2x^2}$

$2\sec(x^2)\tan(x^2)+\frac{2}{x}$

$2x\tan(x^2)+\frac{2}{x}$

$2x\sec(x^2)\tan(x^2)+\frac{2}{x}$

Correct answer:

$2x\sec(x^2)\tan(x^2)+\frac{2}{x}$

Explanation:

The derivative of the function is equal to

$f'=2x\sec(x^2)\tan(x^2)+\frac{4x}{2x^2}=2x\sec(x^2)\tan(x^2)+\frac{2}{x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(x)=\frac{1}{x}$

Note that the chain rule was used on the secant function as well as the natural logarithm function.

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Example Question #431 : Ap Calculus Bc

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

xe^y=y^2

Possible Answers:

$\frac{e^y}{2y+2xe^y}$

$\frac{y}{x}$

$\frac{-xe^{2y}}{y}$

$\frac{e^y}{-2y+2xe^y}$

$\frac{e^y}{2y-2xe^y}$

Correct answer:

$\frac{e^y}{2y-2xe^y}$

Explanation:

To determine $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we must take the derivative of both sides of the equation with respect to x:

$e^y+2xe^y \frac{\mathrm{d} y}{\mathrm{d} x}= 2y\frac{\mathrm{d} y}{\mathrm{d} x}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

Rearranging and solving for $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{e^y}{2y-2xe^y}$

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Example Question #61 : Computation Of Derivatives

A curve in the xy plane is given implictly by

xy^2+x^2y-x=5 .

Calculate the slope of the line tangent to the curve at the point (1, -3) .

Possible Answers:

$-\frac{16}{5}$

$\frac{14}{5}$

$-\frac{3}{5}$

$\frac{2}{5}$

$-\frac{4}{5}$

Correct answer:

$\frac{2}{5}$

Explanation:

Differentiate both sides with respect to using the chain rule and the product rule as:

$\left ((1)(y^2) + (x)(2y)\left (\frac{dy}{dx} \right ) \right )+\left ((2x)(y)+(x^2)\left(\frac{dy}{dx} \right ) \right )-1=0$

Then solve for dy/dx as if it were our unknown:

$\left (\frac{dy}{dx} \right )(2xy+x^2)=1-2xy-y^2$

$\frac{dy}{dx}=\frac{1-2xy-y^2}{2xy+x^2}$ .

Finally, evaluate dy/dx at the point (1,-3) to obtain the slope through that point:

$\frac{1-2(1)(-3)-(-3)^2}{2(1)(-3)+(1)^2} = \frac{-2}{-5}=\frac{2}{5}$ .

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Example Question #62 : Computation Of Derivatives

Screen shot 2016 03 31 at 11.11.35 am

Figure. Squircle of "radius" 1

A squircle is a curve in the xy plane that appears like a rounded square, but whose points satisfy the following equation (analogous to the Pythagorean theorem for a circle)

x^4+y^4=R^4

where the constant is the "radius" of the squircle.

Using implicit differentiation, obtain an expression for $\frac{dy}{dx}$ as a function of both and .

Possible Answers:

$\frac{4x^3}{y^3}$

$-\frac{x^3}{y^3}$

$-\frac{4x^3}{y^3}$

$-\frac{x^4}{4y^3}$

$\frac{x^3}{y^3}$

Correct answer:

$-\frac{x^3}{y^3}$

Explanation:

Differentiate both sides of the equation with respect to , using the chain rule on the y^4 term:

x^4+y^4=R^4

$4x^3+4y^3\left (\frac{dy}{dx} \right )=0$

Then solve for $\frac{dy}{dx}$ as if it were our unknown:

$\frac{dy}{dx}=\frac{-4x^3}{4y^3}=-\frac{x^3}{y^3}$ .

Comparing this to the figure, our answer makes sense, because the slope of the squircle is wherever x=0 (as it crosses the axis) and undefined (vertical) wherever y=0 (as it crosses the axis). Lastly, we note that in the first quadrant (where x>0 and y>0 ), the slope of the squircle is negative, which is exactly what we observe in the figure.

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AP Calculus BC : Chain Rule and Implicit Differentiation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #51 : Computation Of Derivatives

Example Question #61 : Computation Of Derivatives

Example Question #161 : Derivatives

Example Question #431 : Ap Calculus Bc

Example Question #61 : Computation Of Derivatives

Example Question #62 : Computation Of Derivatives

All AP Calculus BC Resources

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