Chain Rule and Implicit Differentiation - AP Calculus BC

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Question

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

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

Answer

To find , 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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Question

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

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

Answer

To find , 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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Question

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

Evaluate .

Answer

To find , 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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Question

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

Evaluate .

Answer

To find , substitute and use the chain rule:

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

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

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

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

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

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

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

So

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

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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Question

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

Evaluate .

Answer

To find , 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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Question

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

Answer

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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Question

Find dy/dx by implicit differentiation:

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

Answer

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:

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

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

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

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Question

Find dx/dy by implicit differentiation:

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

Answer

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:

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

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

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

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Question

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

Answer

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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Question

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

Answer

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

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Question

$\frac{d}{dx } \cos(\ln(x))$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, $f(x) = \cos(x)$ and $g(x) = \ln(x)$ . The derivative is $- \sin (\ln x ) \cdot \frac{1}{x } = \frac{- \sin (\ln x )}{x }$ .

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Question

$\frac{d}{dx } \tan ^ 3 x$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, and $g(x) = \tan x$ . The derivative is $3 (\tan x) ^2 \cdot( 1 + \tan ^2 x ) = 3 \tan ^2 x (1 + \tan ^2 x ) = 3 \tan ^2 x + 3 \tan ^4 x$

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Question

$\frac{d}{dx } 4(x-2)^2$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, and . and .

The derivative is $8(x-2) \cdot 1 = 8x -16$

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Question

$\frac{d}{dx} e^{\sin x}$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, and $g(x) = \sin x$ . Since and $g'(x) = \cos x$ , the derivative is $e ^ {\sin x } \cdot \cos x$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x} 3 ^ {x^2 }$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, and . Since $f'(x) = 3^ {x} \cdot \ln(3)$ and , the derivative is $3^{x^2 } \cdot \ln(3 ) \cdot 2x$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x} \enspace \frac{1}{4} \cdot 2^{4x-1}$

Answer

According to the chain rule, $f(g(x)) ' = f'(g(x)) \cdot g'(x)$ . In this case, $f(x) = \frac{1}{4} \cdot 2^x$ and . Here $f'(x) = \frac{1}{4} \cdot 2^x \cdot \ln(2)$ and . The derivative is:

$\frac{1}{4} \cdot 2 ^{4x-1} \cdot \ln(2) \cdot 4 = 2^{4x-1} \cdot \ln(2)$

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Question

Given the relation , find $\frac{dy}{dx}$ .

Answer

We begin by taking the derivative of both sides of the equation.

$\frac{d}{dx}(y^2)=\frac{d}{dx}(x^2+y)$ .

. (The left hand side uses the Chain Rule.)

.

$y' = \frac{2x}{2y-1}$

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

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Question

Given the relation , find .

Answer

We can use implicit differentiation to find . We being by taking the derivative of both sides of the equation.

$\frac{d}{dx}(x^2+xy^2)=\frac{d}{dx}(y)$

$\frac{d}{dx}(x^2)+\frac{d}{dx}(xy^2)=y'$

$2x +[x \times 2y' + y^2 \times 1] = y'$ (This line uses the product rule for the derivative of .)

$\frac{2x+y^2}{-2x+1} = y'$

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Question

If $f(x) = e^{e^x}$ , find .

Answer

Since we have a function inside of a another function, the chain rule is appropriate here.

The chain rule formula is

$(h(g(x)))' = h'(g(x))\times g'(x)$ .

In our function, both are

So we have

$f'(x) = (h(g(x)))' = e^{e^x} \times e^x = e^{e^x+1}$

and

$f'(1)=e^{e+1}$ .

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Question

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ of the following:

Answer

To find $\frac{\mathrm{d} y}{\mathrm{d} x}$ we must use implicit differentiation, which is an application of the chain rule.
Taking $\frac{\mathrm{d} }{\mathrm{d} x}$ of both sides of the equation, we get

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

using the following rules:

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

Note that for every derivative with respect to x of a function with y, the additional term dy/dx appears; this is because of the chain rule, where y=g(x), so to speak, for the function it appears in.

Using algebra, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-e^y-2x}{xe^y+2y-1}$

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