Chain Rule and Implicit Differentiation - AP Calculus BC

Card 0 of 260

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

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

Tap to see back →

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}$

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

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

Tap to see back →

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$

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

Evaluate .

Tap to see back →

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:

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

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

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

Evaluate .

Tap to see back →

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}$$

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

Evaluate .

Tap to see back →

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$}$

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

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

Tap to see back →

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)}$=2/2=1$ .

Find dy/dx by implicit differentiation:

Tap to see back →

To find dy/dx we must take the derivative of the given function implicitly. Notice the term 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:

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:

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

This is one of the answer choices.

Find dx/dy by implicit differentiation:

Tap to see back →

To find dx/dy we must take the derivative of the given function implicitly. Notice the term 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:

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:

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

This is one of the answer choices.

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

Tap to see back →

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

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

Tap to see back →

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 $\cos(x^2$)$

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

Tap to see back →

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 }$$ .

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

Tap to see back →

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 $3 (\tan x) ^2 \cdot( 1 + \tan ^2 x ) = 3 \tan ^2 x (1 + \tan ^2 x ) = 3 \tan ^2 x + 3 \tan ^4 x$

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

Tap to see back →

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

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

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

Tap to see back →

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 $e ^ {\sin x } \cdot \cos x$

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

Tap to see back →

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

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

Tap to see back →

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)$

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

Tap to see back →

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

.

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

.

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

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

Given the relation , find .

Tap to see back →

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

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

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

Tap to see back →

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

.

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

Tap to see back →

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$+2y-1}$