Chain Rule

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We compute the derivative of a composition.

The Chain Rule

If $f(x)$ and $g(x)$ are differentiable functions and $h(x) = f(g(x))$ , then $h'(x) = f'(g(x))g'(x).$ This important differentiation rule can also be written as $f(g(x))' = f'(g(x))g'(x)$ or, using Leibniz notation, as $\frac {d}{dx}f(g(x)) = f'(g(x))g'(x).$

In words, the derivative of a composition is the derivative of the outside, with the inside left in, times the derivative of the inside. Or, we can shorten the phrasing slightly to simply ‘the derivative of the outside times the derivative of the inside’.

Find $h'(x)$ if $h(x) = \sin (2x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 2x \quad \text {and} \quad f(x) = \sin (x).$

To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 2; \quad \text {next}$ $f'(x) = \cos (x), \quad \text {so}$ $f'(g(x)) =\cos (g(x)) = \cos (2x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \cos (2x) \cdot 2 = 2\cos (2x).$

Here is a video of Example 1

Compute $\frac {d}{dx} \sin (5x)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The ‘‘outside’’ function is $f(x) = \sin (x)$ and the ‘‘inside’’ function is $g(x) = 5x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\sin (5x)$ with respect to $x$ is $\answer [given]{5\cos (5x)}$

Find $h'(x)$ if $h(x) = e^{5x}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 5x \quad \text {and} \quad f(x) = e^x.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First: $g'(x) = 5; \quad \text {next}$ $f'(x) = e^x, \quad \text {so}$ $f'(g(x)) =e^{g(x)} = e^{5x}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = e^{5x} \cdot 5 = 5e^{5x}.$

Here is a video of the example

Compute $\frac {d}{dx} e^{3x}$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = e^x$ and the inside function is $g(x) = 3x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $e^{3x}$ with respect to $x$ is $\answer [given]{3e^{3x}}$

Find $h'(x)$ if $h(x) = \cos (x^2 + 1)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^2 + 1 \quad \text {and} \quad f(x) = \cos (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 2x; \quad \text {next}$ $f'(x) = -\sin (x), \quad \text {so}$ $f'(g(x))= -\sin (g(x)) = -\sin (x^2 + 1).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = -\sin (x^2 + 1) \cdot 2x = -2x\sin (x^2 + 1).$

Here is a video of the example

Compute $\frac {d}{dx} \cos (2-x^3)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \cos (x)$ and the inside function is $g(x) = 2-x^3$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\cos (2-x^3)$ with respect to $x$ is $\answer [given]{3x^2\sin (2-x^3)}$

Find $h'(x)$ if $h(x) = (x^3 - 4x + 2)^7$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^3 - 4x + 2 \quad \text {and} \quad f(x) = x^7.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) =3x^2 - 4; \quad \text {next}$ $f'(x) = 7x^6 , \quad \text {so}$ $f'(g(x)) = 7[g(x)]^6 = 7(x^3 - 4x + 2)^6 .$ By the chain rule, $h'(x) = f'(g(x))g'(x) = 7(x^3 + 4x + 2)^6(3x^2 - 4)=7(3x^2 - 4)(x^3 - 4x + 2)^6.$

Here is a video of the example

Compute $\frac {d}{dx} (x^3 - 4x + 5)^8$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = x^8$ and the inside function is $x^3 - 4x + 5$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $(x^3 - 4x + 5)^8$ with respect to $x$ is $\answer [given]{8(3x^2 - 4)(x^3 - 4x + 5)^7 }$

Find $h'(x)$ if $h(x) = \sqrt {x^2 + 4}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^2 + 4 \quad \text {and} \quad f(x) = \sqrt x.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 2x; \quad \text {next}$ $f'(x) = \frac {1}{2\sqrt x} , \quad \text {so}$ $f'(g(x)) = \frac {1}{2\sqrt {g(x)}}=\frac {1}{2\sqrt {x^2 + 4}}.$ By the chain rule,

$\begin{align*} h'(x) &= f'(g(x))g'(x)\\ &= \frac{1}{2\sqrt {x^2 + 4}}\cdot 2x\\ &= \frac{2x}{2\sqrt {x^2 + 4}}\\ =\frac{x}{\sqrt {x^2 + 4}}. \end{align*}$

Here is a video of the example

Compute $\frac {d}{dx} \sqrt {x^2 - 3}$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \sqrt {x}$ and the inside function is $g(x) = x^2 - 3$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\sqrt {x^2 - 3}$ with respect to $x$ is $\answer [given]{\frac {x}{\sqrt {x^2 - 3}}}$

Find $h'(x)$ if $h(x) = \tan (3x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 3x \quad \text {and} \quad f(x) =\tan (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) =3; \quad \text {next}$ $f'(x) =\sec ^2(x) , \quad \text {so}$ $f'(g(x)) = \sec ^2(g(x)) = \sec ^2(3x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \sec ^2(3x) \cdot 3= 3\sec ^2(3x).$

Here is a video of the example

Compute $\frac {d}{dx} \tan (x^4)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \tan (x)$ and the inside function is $g(x) = x^4$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\tan (x^4)$ with respect to $x$ is $\answer [given]{4x^3 \sec ^2(x^4)}$

What is the derivative of $\sec (2x)$ ?

$\sec (2x)\tan (2x)$ $2\sec (x)\tan (x)$ $-2\sec (2x)\tan (2x)$ $2\sec (2x)\tan (2x)$

Find $h'(x)$ if $h(x) = \frac {3}{(2x + 5)^4}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 2x+ 5 \quad \text {and} \quad f(x) = \frac {3}{x^4} = 3x^{-4}.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 2; \quad \text {next}$ $f'(x) = -12x^{-5}, \quad \text {so}$ $f'(g(x)) = -12[g(x)]^{-5} = -12(2x+5)^{-5} = -\frac {12}{(2x+5)^5}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = -\frac {12}{(2x+5)^5}\cdot 2 = -\frac {24}{(2x+5)^5}.$

Here is a video of the example

Compute $\frac {d}{dx} \frac {2}{(3x - 8)^5}$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = 2x^{-5}$ and the inside function is $g(x) = 3x - 8$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\frac {2}{(3x - 8)^5}$ with respect to $x$ is $\answer [given]{\frac {-30}{(3x - 8)^6}}$

Find $h'(x)$ if $h(x) = \sin (x^3)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^3 \quad \text {and} \quad f(x) = \sin (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 3x^2; \quad \text {next}$ $f'(x) =\cos (x) , \quad \text {so}$ $f'(g(x)) = \cos (g(x)) = \cos (x^3).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \cos (x^3)\cdot 3x^2 = 3x^2\cos (x^3).$

Here is a video of the example

Compute $\frac {d}{dx} \sin (3x^5)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \sin (x)$ and the inside function is $g(x) = 3x^5$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\sin (3x^5)$ with respect to $x$ is $\answer [given]{\cos (3x^5)\cdot 15x^4}$

Find $h'(x)$ if $h(x) = \sin ^3(x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) =\sin (x) \quad \text {and} \quad f(x) = x^3.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) =\cos (x); \quad \text {next}$ $f'(x) = 3x^2, \quad \text {so}$ $f'(g(x)) = 3g^2(x) = 3\sin ^2(x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = 3\sin ^2(x)\cos (x).$

Here is a video of the example

Compute $\frac {d}{dx} \sin ^4(x)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = x^4$ and the inside function is $g(x) = \sin (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\sin ^4(x)$ with respect to $x$ is $\answer [given]{4\sin ^3(x)\cos (x)}$

Find $h'(x)$ if $h(x) = e^{-x^2}$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = -x^2 \quad \text {and} \quad f(x) =e^x.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = -2x; \quad \text {next}$ $f'(x) = e^x, \quad \text {so}$ $f'(g(x)) = e^{g(x)} = e^{-x^2}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = e^{-x^2} (-2x) \quad \text {or} \quad -2xe^{-x^2}.$

Here is a video of the example

Compute $\frac {d}{dx} e^{\sqrt x}$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = e^x$ and the inside function is $g(x) = \sqrt x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $e^{\sqrt x}$ with respect to $x$ is $\answer [given]{\frac {e^{\sqrt x}}{2\sqrt x}}$

Find $h'(x)$ if $h(x) = \ln (x^4 +1)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = x^4 + 1 \quad \text {and} \quad f(x) = \ln (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 4x^3; \quad \text {next}$ $f'(x) = \frac {1}{x}, \quad \text {so}$ $f'(g(x)) =\frac {1}{g(x)} = \frac {1}{x^4 + 1} .$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \frac {1}{x^4 + 1} \cdot 4x^3 = \frac {4x^3}{x^4 + 1}.$

Here is a video of the example

Compute $\frac {d}{dx} \ln (x^3 -2)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \ln (x)$ and the inside function is $g(x) = x^3 - 2$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\ln (x^3 - 2)$ with respect to $x$ is $\answer [given]{\frac {3x^2}{x^3 - 2}}$

Find $h'(x)$ if $h(x) = \ln ^5(x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = \ln (x) \quad \text {and} \quad f(x) = x^5.$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = \frac {1}{x}; \quad \text {next}$ $f'(x) = 5x^4, \quad \text {so}$ $f'(g(x)) = 5g^4(x) = 5\ln ^4(x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = 5\ln ^4(x)\cdot \frac {1}{x} = \frac {5\ln ^4(x)}{x}.$

Here is a video of the example

Compute $\frac {d}{dx} \ln ^3(x)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = x^3$ and the inside function is $g(x) = \ln (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\ln ^3(x)$ with respect to $x$ is $\answer [given]{\frac {3\ln ^2(x)}{x}}$

Find $h'(x)$ if $h(x) = \ln (\sin (x))$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = \sin (x) \quad \text {and} \quad f(x) = \ln (x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = \cos (x); \quad \text {next}$ $f'(x) = \frac {1}{x}, \quad \text {so}$ $f'(g(x)) = \frac {1}{g(x)} = \frac {1}{\sin (x)}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \frac {1}{\sin (x)} \cdot \cos (x) = \frac {\cos (x)}{\sin (x)} = \cot (x).$

Here is a video of the example

Compute $\frac {d}{dx} \ln (\sec (x))$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \ln (x)$ and the inside function is $g(x) = \sec (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\ln (\sec (x))$ with respect to $x$ is $\answer [given]{\tan (x)}$

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

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \cos (x)$ and the inside function is $g(x) = \ln (x)$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\cos (\ln (x))$ with respect to $x$ is $\answer [given]{\frac {-\sin (\ln (x))}{x}}$

Find $h'(x)$ if $h(x) = \tan ^{-1}(4x)$ .
We write $h(x)$ as a composition: $h(x)=f(g(x))$ with $g(x) = 4x \quad \text {and} \quad f(x) = \tan ^{-1}(x).$ To find $h'(x)$ we need $f'(g(x))$ and $g'(x)$ . First, $g'(x) = 4; \quad \text {next}$ $f'(x) = \frac {1}{1+x^2} , \quad \text {so}$ $f'(g(x)) = \frac {1}{1+g^2(x)} =\frac {1}{1+(4x)^2}=\frac {1}{1+16x^2}.$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \frac {1}{1+16x^2} \cdot 4 = \frac {4}{1+16x^2}.$

Here is a video of the example

Compute $\frac {d}{dx} \tan ^{-1}(3x)$

The chain rule says: $\frac {d}{dx} f(g(x)) = f'(g(x))\cdot g'(x)$

The outside function is $f(x) = \tan ^{-1}(x)$ and the inside function is $g(x) = 3x$ .

Leave the inside in, $f'(g(x))$

Multiply by the derivative of the inside, $g'(x)$

The derivative of $\tan ^{-1}(3x)$ with respect to $x$ is $\answer [given]{\frac {3}{1+(3x)^2}}$

Here are some detailed, lecture style videos on the chain rule: