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AP Calculus BC : Chain Rule and Implicit Differentiation

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Example Questions

Example Question #1 : Chain Rule And Implicit Differentiation

Find dy/dx by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Possible Answers:

$\frac{3x^2-\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{e^y+\tan^{-1}(y)-x^3}{x}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{e^y+3x^2}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

Correct answer:

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Explanation:

To find dy/dx we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dx}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dx}))+(1)(\frac{dx}{dx})(\tan^{-1}(y))]=e^y\frac{dy}{dx}$

Notice that anytime we take the derivative of a term with x involved we place a "dx/dx" next to it, but this is equal to "1".

So this now becomes:
$3x^2+[\frac{x}{1+y^2}\frac{dy}{dx}+\tan^{-1}(y)}]=e^y\frac{dy}{dx}$

Now if we place all the terms with a "dy/dx" onto one side and factor out we can solved for it:

$3x^2+\tan^{-1}(y)=e^y\frac{dy}{dx}-\frac{x}{1+y^2}\frac{dy}{dx}$

$3x^2+\tan^{-1}(y)=\frac{dy}{dx}(e^y-\frac{x}{1+y^2})$

$\frac{dy}{dx}=\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

This is one of the answer choices.

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Example Question #2 : Chain Rule And Implicit Differentiation

Find dx/dy by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Possible Answers:

$\frac{3x^2+e^y}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{\tan^{-1}(y)-\frac{x}{1+y^2}}{3x^2+e^y}$

$\frac{\frac{x}{1+y^2}+e^y}{5x^2-\tan^{-1}(y)}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Correct answer:

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

Explanation:

To find dx/dy we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

Product Rule: $\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dy}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dy}))+(1)(\frac{dx}{dy})(\tan^{-1}(y))]=e^y\frac{dy}{dy}$

Notice that anytime we take the derivative of a term with y involved we place a "dy/dy" next to it, but this is equal to "1".

So this now becomes:
$3x^2\frac{dx}{dy}+[\frac{x}{1+y^2}+\tan^{-1}(y)\frac{dx}{dy}]=e^y$

Now if we place all the terms with a "dx/dy" onto one side and factor out we can solved for it:

$3x^2\frac{dx}{dy}+\tan^{-1}(y)\frac{dx}{dy}=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}[3x^2+\tan^{-1}(y)]=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}=\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

This is one of the answer choices.

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Example Question #3 : Chain Rule And Implicit Differentiation

Use implicit differentiation to find the slope of the tangent line to $\small y^3-lny=x^2$ at the point $\small (1,1)$ .

Possible Answers:

$\small 1$

$\small -2$

$\small 0$

$\small -1$

$\small 2$

Correct answer:

$\small 1$

Explanation:

We must take the derivative $\small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get

$\small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$ , and on the right side we'll get $\small 2x$ .

We include the $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\small y$ is a function of $\small x$ , so its derivative is unknown (hence we are trying to solve for it!).

Now we can factor out a $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\small 3y^2-1/y$ in order to solve for $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ .

Doing this gives you

$\small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$ .

We want to find the slope at $\small (1,1)$ , so we can sub in $\small 1$ for $\small x$ and $\small y$ .

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$ .

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Example Question #4 : Chain Rule And Implicit Differentiation

$g(x) = e ^{x^{2}+x}$

Evaluate g '(3 ) .

Possible Answers:

$7 e^{12}$

$12 e^{12}$

$e^{7}$

$7 e^{9}$

$12 e^{11}$

Correct answer:

$7 e^{12}$

Explanation:

To find g'(x) , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

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Example Question #5 : Chain Rule And Implicit Differentiation

$g (x) = \frac{1}{4x - 7}$

Evaluate g ' (1 ) .

Possible Answers:

$-\frac{4}{3}$

$\frac{4}{3}$

$\frac{4}{9}$

$- \frac{4}{9}$

Undefined

Correct answer:

$- \frac{4}{9}$

Explanation:

To find g'(x) , substitute u =4x - 7 and use the chain rule:

$g (x) = \frac{1}{4x - 7}$

$g '(x) = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{4x - 7}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{u}$

$= \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \frac{1}{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x} (4x-7) \cdot \frac{\mathrm{d} }{\mathrm{d} u} u ^{-1}$

$=4 \cdot (-1 ) u ^{-1-1}$

$=-4 u ^{-2} = - \frac{4}{u^{2}}= - \frac{4}{(4x-7)^{2}}$

$g'(x)= - \frac{4}{(4x-7)^{2}}$

and

$g'(1)= - \frac{4}{(4 \cdot 1 -7)^{2}}$

$= - \frac{4}{(4 -7)^{2}}$

$= - \frac{4}{(-3)^{2}} = - \frac{4}{9}$

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Example Question #6 : Chain Rule And Implicit Differentiation

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

Evaluate $g' \left ( \pi )$ .

Possible Answers:

$-\sqrt{2}$

Undefined

$\sqrt{2}$

Correct answer:

Explanation:

To find g'(x) , substitute $u = x +\frac{ \pi }{4}$ and use the chain rule:

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right )\cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right ) \cdot \sec ^{2} u$

$=1 \cdot \sec ^{2}\left ( x +\frac{ \pi }{4}\right ) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(x) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\left ( \pi +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\frac{ 5 \pi }{4}$

$= \left ( - \sqrt{2 }\right )^{2} = 2$

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Example Question #7 : Chain Rule And Implicit Differentiation

$g (x) = 3 ^{x^{2}}$

Evaluate g ' (-1 ) .

Possible Answers:

$-\frac{3 \ln 3}{2}$

$-6 \ln 3$

Correct answer:

$-6 \ln 3$

Explanation:

To find g'(x) , substitute $u = x ^{2}$ and use the chain rule:

$g (x) = 3 ^{x^{2}}$

$g ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{x^{2}}$

$= \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= \frac{\mathrm{d}}{\mathrm{d} x} x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= 2x \cdot \ln 3 \cdot 3 ^{u}$

$= 2x \ln 3 \cdot 3 ^{x^2}$

So $g'(x)= 2x \ln 3 \cdot 3 ^{x^2}$

and

$g'(-1)= 2(-1) \ln 3 \cdot 3 ^{(-1)^2}$

$= -2 \ln 3 \cdot 3 ^{1} = -2 \ln 3 \cdot 3 = -6 \ln 3$

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Example Question #8 : Chain Rule And Implicit Differentiation

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

Evaluate $g ' \left ( \frac{1}{2} \right )$ .

Possible Answers:

$- \pi \sqrt{2}$

$- \frac{ \pi \sqrt{2}}{2}$

$\frac{ \pi \sqrt{2}}{2}$

$- \frac{ \sqrt{2}}{2 \pi}$

$\pi \sqrt{2}$

Correct answer:

$- \frac{ \pi \sqrt{2}}{2}$

Explanation:

To find g'(x) , substitute $u = \pi x^{2}$ and use the chain rule:

$g \left (x \right )= \cos \left ( \pi x^{2}\right )$

$g'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos \left ( \pi x^{2}\right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} \cos u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \pi x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$= \pi \cdot 2 x\cdot \left ( - \sin u \right )$

$= \pi \cdot 2 x\cdot \left [- \sin \left ( \pi x^{2} ) \right ]$

$=-2 \pi x \sin \left ( \pi x^{2} )$

So $g'(x)=-2 \pi x \sin \left ( \pi x^{2} )$

and

$g ' \left ( \frac{1}{2} \right )=-2 \pi\cdot \frac{1}{2} \sin \left [\pi \cdot \left ( \frac{1}{2} \right )^{2} \right ]$

$=- \pi \sin \left ( \frac{\pi}{4} \right )$

$=- \pi \cdot \frac{\sqrt{2}}{2} = - \frac{ \pi \sqrt{2}}{2}$

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Example Question #411 : Ap Calculus Bc

$\frac{d}{dx} \ln(3x-7)$

Possible Answers:

$\ln(3)$

$\frac{1}{3x-7}$

$\frac{3}{3x-7}$

$3 \ln (3x-7)$

Correct answer:

$\frac{3}{3x-7}$

Explanation:

Consider this function a composition of two functions, f(g(x)). In this case, f(x) is ln(x) and g(x) is 3x - 7. The derivative of ln(x) is 1/x, and the derivative of 3x - 7 is 3. The derivative is then $\frac{1}{3x-7} \cdot 3 = \frac{3}{3x-7}$ .

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Example Question #1 : Chain Rule And Implicit Differentiation

$\frac{d}{dx } 2 \sin (x^2 )$

Possible Answers:

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

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

$-4x \cdot \cos(x^2)$

$4x \cdot \cos(x^2)$

Correct answer:

$4x \cdot \cos(x^2)$

Explanation:

Consider this function a composition of two functions, f(g(x)). In this case, $f(x) =2 \sin (x)$ and g(x) = x^2 . According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . Here, $f'(g(x)) = 2 \cos (x^2 )$ and g'(x) = 2x , so the derivative is $2 \cos (x^2 ) \cdot 2x = 4x \cdot \cos(x^2)$

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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 #1 : Chain Rule And Implicit Differentiation

Find dy/dx by implicit differentiation:

Example Question #2 : Chain Rule And Implicit Differentiation

Find dx/dy by implicit differentiation:

Example Question #3 : Chain Rule And Implicit Differentiation

Example Question #4 : Chain Rule And Implicit Differentiation

Example Question #5 : Chain Rule And Implicit Differentiation

Example Question #6 : Chain Rule And Implicit Differentiation

Example Question #7 : Chain Rule And Implicit Differentiation

Example Question #8 : Chain Rule And Implicit Differentiation

Example Question #411 : Ap Calculus Bc

Example Question #1 : Chain Rule And Implicit Differentiation

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