2.7 Chain Rule

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We compute the derivative of a composition.

1 Compositions

Composition is an operation that is unique to functions. It involves taking the output of one function and using it as the input of another function. The composition of two functions $f$ and $g$ is the function $f \circ g$ defined by $(f \circ g)(x) = f(g(x))$ We refer to $g$ as the “inside” function and $f$ as the “outside” function.

example 1 Identify the inside and the outside functions in these compositions:

(a): The function $\sqrt {x^2 + 9}$ is the composition of the inside function $g(x) =x^2 + 9$ and the outside function $f(x) = \sqrt {x}$ .
(b): The function $e^{-x^2}$ is the composition of the inside function $g(x) = -x^2$ and the outside function $f(x) = e^x$ .
(c): The function $\tan ^3(x)$ is the composition of the inside function $g(x) = \tan (x)$ and the outside function $f(x) = x^3$ .
Note that the inside function is the one that acts first on $x$ and the outside function is the one that acts last on $x$ .

(problem 1) Identify the inside and the outside functions in these compositions:

$(x^3 + 1)^5$ Inside function, $g(x) = \answer {x^3 + 1}$ and outside function, $f(x) = \answer {x^5}$
$\ln (\cos (x))$ Inside function, $g(x) = \answer {\cos (x)}$ and outside function, $f(x) = \answer {\ln (x)}$
$e^{\sqrt {x}}$ Inside function, $g(x) = \answer {\sqrt {x}}$ and outside function, $f(x) = \answer {e^x}$

A composition can be written as a chain of variables: if $y = f(g(x))$ , then we can introduce a new variable, $u$ and we can write $y = f(u)$ where $u = g(x)$ .

example 2 Write the following compositions as a chain of variables:

(a): The function $y = \sqrt {x^2 + 9}$ can be written as $y = \sqrt {u}$ where $u = x^2 + 9$ .
(b): The function $y =e^{-x^2}$ can be written as $y = e^u$ where $u = -x^2$ .
(c): The function $\tan ^3(x)$ can be written as $y = u^3$ where $u = \tan (x)$ .

(problem 2) Write the following compositions as a chain of variables:

(a): The single equation $y = (x^3 + 1)^5$ can be written as $y = f(u) = \answer {u^5}$ where $u = g(x) = \answer {x^3 + 1}$
(b): The single equation $y = \ln (\cos (x))$ ; can be written as $y = f(u) = \answer {\ln (u)}$ where $u = g(x) = \answer {\cos (x)}$
(c): The single equation $y = e^{\sqrt {x}}$ ; can be written as $y = f(u) = \answer {e^u}$ where $u = g(x) = \answer {\sqrt {x}}$

2 The Chain Rule in Prime Notation

The chain rule is the formula for computing the derivative of a composition.

Chain Rule If $f(x)$ and $g(x)$ are differentiable functions, then $\frac {d}{dx}\left [f(g(x))\right ] = f'(g(x))g'(x).$

The formula can be expressed by saying that the derivative of a composition is “the derivative of the outside times the derivative of the inside.” It must be noted that the derivative of the outside function is evaluated at the inside function.

example 3 Find $\dfrac {d}{dx}\left [\sin (x^4)\right ]$ .
The function $\sin (x^4)$ is a composition with $g(x) = x^4$ inside and $f(x) = \sin (x)$ outside. To find the derivative of the composite function $\sin (x^4)$ we need to use the chain rule. The chain rule formula requires both $f'(g(x))$ and $g'(x)$ . The second is simpler, so we start there: $g'(x) = 4x^3$ . To find $f'(g(x))$ we first find $f'(x)$ and then substitute $g(x)$ for $x$ . We have, $f'(x) = \cos (x) \qquad \mbox {so,} \qquad f'(g(x)) = \cos (g(x)) = \cos (x^4).$ We can now conclude using the chain rule that $\frac {d}{dx}\left [\sin (x^4)\right ] = f'(g(x))g'(x) = \cos (x^4) \cdot 4x^3 = 4x^3\cos (x^4).$

(problem 3a) Find the indicated derivative: $\frac {d}{dx} \sin (x^3) = \answer {3x^2\cos (x^3)}.$

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 $g(x) = x^3$ .

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

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

Here is a video solution of problem 3a:

(problem 3b) Find the indicated derivative: $\frac {d}{dx} \tan (x^3 - 7) = \answer {3x^2 \sec ^2(x^3 - 7)}.$

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 $g(x) = x^3 - 7$ .

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

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

example 4 Find $\dfrac {d}{dx}\left [e^{-x^2}\right ]$ .
The function $e^{-x^2}$ is a composition with $g(x) = -x^2$ inside and $f(x) = e^x$ outside. To find the derivative of the composite function $e^{-x^2}$ we need to use the chain rule. The chain rule formula requires both $f'(g(x))$ and $g'(x)$ . The second is simpler, so we start there: $g'(x) = -2x$ . To find $f'(g(x))$ we first find $f'(x)$ and then substitute $g(x)$ for $x$ . We have, $f'(x) = e^x \qquad \mbox {so,} \qquad f'(g(x)) = e^{g(x)} = e^{-x^2}$ We can now conclude using the chain rule that $\frac {d}{dx}\left [e^{-x^2}\right ] = f'(g(x))g'(x) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}$

(problem 4a)

Compute $\frac {d}{dx} e^{\sqrt x} = \answer {\frac {e^{\sqrt x}}{2\sqrt x}}.$

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, $g'(x)$

(problem 4b)

Compute $\frac {d}{dx} e^{\cos (x)} = \answer {-e^{\cos (x)}\sin (x)}.$

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 $g(x) = \cos (x)$ .

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

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

example 5 Find the values of $x$ where the tangent line to the graph of $y = (x^4 + x^2 + 1)^4$ is horizontal.
The tangent line is horizontal where the derivative is zero. We write $y$ as a composition: $y = f(g(x))$ with $g(x) = x^4 + x^2 + 1 \quad \text {and} \quad f(x) = x^4.$ To find $y'$ we need $f'(g(x))$ and $g'(x)$ . We have, $g'(x) = 4x^3 + 2x; \quad \text {next}$ $f'(x) = 4x^3, \quad \text {so}$ $f'(g(x)) = 4[g(x)]^3 = 4(x^4 + x^2 + 1)^3.$ By the chain rule, $y' = f'(g(x))g'(x) = 4(x^4 + x^2 + 1)^3(4x^3 + 2x).$ The derivative is zero when $4(x^4 + x^2 + 1)^3(4x^3 + 2x) = 0$ . Since $4(x^4 + x^2 + 1)^3$ is never zero, we need to solve $4x^3 + 2x = 0$ . We can factor out $2x$ to get $2x(2x^2 + 1) = 0$ . The only solution is $x = 0$ .

(problem 5) Find the values of $x$ where the tangent line to the graph of $y = (x^2 - 5x + 4)^5$ is horizontal.
First we find the derivative: $y' = \answer {5(x^2 - 5x + 4)^4(2x - 5)}$

The inside function is $g(x) = x^2 - 5x + 4$ and the outside function is $f(x) = x^5$ .

The tangent line is horizontal where the derivative is zero.

In increasing order, the $x$ -values where the tangent line is horizontal are $x = \answer {1}$ , $\answer {5/2}$ , $x = \answer {4}$ .

3 Leibniz Notation for the Chain Rule

The chain rule can also be expressed in Leibniz notation. A composition $y = f(g(x))$ can be written using a chain of variables as: $y = f(u)$ where $u = g(x)$ . Observe that $\frac {dy}{du} = f'(u) = f'(g(x))$ and $\frac {du}{dx} = g'(x)$ . Thus, the chain rule in Leibniz notation is given by the elegant formula: $\frac {dy}{dx} = \frac {dy}{du} \cdot \frac {du}{dx},$ In this format, the chain rule expresses the fact that “rates of change multiply.” To see this, consider a small change in the variable $x$ . This produces a change in $u$ which in turn produces a change in $y$ . The chain rule states that the rate of change of $y$ with respect to $x$ is the rate of change of $u$ with respect to $x$ times the rate of change of $y$ with respect to $u$ . For example, if $u$ is changing three times faster than $x$ and $y$ is changing four times faster than $u$ , then $y$ is changing $3 \cdot 4 = 12$ times faster than $x$ .

example 6 Find $\frac {dy}{dx}$ if $y = \cos (x^3)$ .
We unravel the composition by substituting $u$ for $x^3$ leading to the following chain of variables: $y = \cos (u), \text { where } \, u = x^3.$ Then the derivatives are $\frac {dy}{du} = -\sin (u) \,\,\,\text { and } \,\,\,\frac {du}{dx} = 3x^2.$ Finally, the chain rule says to multiply these derivatives: $\frac {dy}{dx} = \frac {dy}{du} \cdot \frac {du}{dx} = -\sin (u) \cdot 3x^2 = -3x^2\sin (x^3).$

(problem 6) Find $\frac {dy}{dx}$ if $y = \ln (x^3 +x^2)$ and simplify the answer.
We unravel the composition by substituting $u$ for $x^3 + x^2$ leading to the following chain of variables: $y = \ln (u), \text { where } \, u = x^3 + x^2$ Then the derivatives are $\frac {dy}{du} = \frac {1}{u} \,\,\,\text { and } \,\,\,\frac {du}{dx} = 3x^2 + 2x.$ Finally, the chain rule says to multiply these derivatives: $\frac {dy}{dx} = \frac {dy}{du} \cdot \frac {du}{dx} = \frac {1}{u} \cdot (3x^2 + 2x) = \frac {3x^2 + 2x}{x^3 + x^2}$ which can be simplified to $\frac {3x + 2}{x^2 + x}$ .

example 7 Find the second derivative of $\sec ^2(x)$ .
We unravel the composition by substituting $u$ for $\sec (x)$ leading to the following chain of variables: $y = u^2, \text { where } \, u = \sec (x).$ Then the derivatives are $\frac {dy}{du} = 2u \,\,\,\text { and } \,\,\,\frac {du}{dx} = \sec (x)\tan (x).$ Finally, the chain rule says to multiply these derivatives: $\frac {d}{dx}\left [\sec ^2(x)\right ] = \frac {dy}{du} \cdot \frac {du}{dx} = 2u \cdot \sec (x)\tan (x) = 2\sec (x) \cdot \sec (x)\tan (x) = 2\sec ^2(x)\tan (x).$ Note that the function $\sec ^2(x)$ that we differentiated in this example is already the derivative of $\tan (x)$ . Hence, we found that the second derivative of $\tan (x)$ is $2\sec ^2(x)\tan (x)$ .

(problem 7) Find the derivative of $\ln ^4(x)$ .
We unravel the composition by substituting $u$ for $\ln (x)$ leading to the following chain of variables: $y = u^4, \text { where } \, u = \ln (x).$ Then the derivatives are $\frac {dy}{du} = 4u \,\,\,\text { and } \,\,\,\frac {du}{dx} = \frac {1}{x}.$ Finally, the chain rule says to multiply these derivatives: $\frac {d}{dx}\left [\ln ^4(x)\right ] = \frac {dy}{du} \cdot \frac {du}{dx} = 4u \cdot \frac {1}{x} = \frac {4\ln (x)}{x}.$

example 8 After $t$ years, the amount of carbon 14 left from a sample of 10 grams is given by the function $C(t) = 10e^{-\frac {t}{5730}}.$ Find the rate at which carbon 14 is decaying after 1000 years.
We write $C(t)$ as a composition: $C(t) = f(g(t))$ with $g(t) = -\frac {t}{5730} \quad \text {and} \quad f(x) = 10e^x.$ To find $C'(t)$ we need $f'(g(t))$ and $g'(t)$ . We have, $g'(t) = -\frac {1}{5730}; \quad \text {next}$ $f'(x) = 10e^x, \quad \text {so}$ $f'(g(t))= 10e^{g(t)} = 10e^{-\frac {t}{5730}}.$ By the chain rule, $C'(t) = f'(g(t))g'(t) = 10e^{-\frac {t}{5730}} \cdot \left (-\frac {1}{5730}\right ) = -\frac {10}{5730}e^{-\frac {t}{5730}}.$ After 1000 years, the rate of decay is $C'(1000) = -\frac {10}{5730}e^{-\frac {1000}{5730}} \approx -0.00175.$ This is a negative value, indicating that the amount of carbon 14 is decreasing over time.

(problem 8) After $t$ million years, the amount of uranium 235 left from a sample of 20 grams is given by the function $U(t) = 20e^{-\frac {5\ln (2)}{3519}t}.$ Find the rate at which uranium 235 is decaying after 2 million years.
We write $U(t)$ as a composition: $U(t) = f(g(t))$ with $g(t) = -\frac {5\ln (2)}{3519}t \quad \text {and} \quad f(x) = 20e^x.$ To find $U'(t)$ we need $f'(g(t))$ and $g'(t)$ . We have, $g'(t) = -\frac {5\ln (2)}{3519}; \quad \text {next}$ $f'(x) = 20e^x, \quad \text {so}$ $f'(g(t))= 20e^{g(t)} = 20e^{-\frac {5\ln (2)}{3519}t}.$ By the chain rule, $U'(t) = f'(g(t))g'(t) = 20e^{-\frac {5\ln (2)}{3519}t} \cdot \left (-\frac {5\ln (2)}{3519}\right ) = -\frac {100\ln (2)}{3519}e^{-\frac {5\ln (2)}{3519}t}.$ After 2 million years, the rate of decay is $U'(2) = -\frac {100\ln (2)}{3519}e^{-\frac {5\ln (2)}{3519}(2)}.$ This is a negative value, indicating that the amount of uranium 235 is decreasing over time.

4 Additional Examples and Problems

example 9 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 $g'(x)$ . We have, $g'(x) = 2x; \quad \text {next}$ $f'(x) = -\sin (x), \quad \text {so}$ $f'(g(x))= -\sin (g(x)) = -\sin (x^2 + 1).$ 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 example 9:

(problem 9) Compute $\frac {d}{dx} \cos (2-x^3) = \answer {3x^2\sin (2-x^3)}.$

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, $g'(x)$

example 10

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 $g'(x)$ . We have, $g'(x) =3x^2 - 4; \quad \text {next}$ $f'(x) = 7x^6 , \quad \text {so}$ $f'(g(x)) = 7[g(x)]^6 = 7(x^3 - 4x + 2)^6 .$ 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 example 10:

(problem 10) Compute $\frac {d}{dx} (x^3 - 4x + 5)^8 = \answer {8(3x^2 - 4)(x^3 - 4x + 5)^7 }.$

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

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

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

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

example 11

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 $g'(x)$ . We have, $g'(x) = 2x; \quad \text {next}$ $f'(x) = \frac {1}{2\sqrt x} , \quad \text {so}$ $f'(g(x)) = \frac {1}{2\sqrt {g(x)}}=\frac {1}{2\sqrt {x^2 + 4}}.$ 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 example 11:

(problem 11) Compute $\frac {d}{dx} \sqrt {x^2 - 3} = \answer {\frac {x}{\sqrt {x^2 - 3}}}.$

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, $g'(x)$

example 12

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 $g'(x)$ . We have, $g'(x) =3; \quad \text {next}$ $f'(x) =\sec ^2(x) , \quad \text {so}$ $f'(g(x)) = \sec ^2(g(x)) = \sec ^2(3x).$ 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 example 12:

(problem 12a) Compute $\frac {d}{dx} \tan (x^4) = \answer {4x^3 \sec ^2(x^4)}.$

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, $g'(x)$

(problem 12b) What is the derivative of $\sec (2x)$ ?

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

example 13 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 $g'(x)$ . We have, $g'(x) = 2; \quad \text {next}$ $f'(x) = -12x^{-5}, \quad \text {so}$ $f'(g(x)) = -12[g(x)]^{-5} = -12(2x+5)^{-5} = -\frac {12}{(2x+5)^5}.$ 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 example 13:

(problem 13) Compute $\frac {d}{dx} \frac {2}{(3x - 8)^5} = \answer {\frac {-30}{(3x - 8)^6}}.$

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, $g'(x)$

example 14 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 $g'(x)$ . We have, $g'(x) = 3x^2; \quad \text {next}$ $f'(x) =\cos (x) , \quad \text {so}$ $f'(g(x)) = \cos (g(x)) = \cos (x^3).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = \cos (x^3)\cdot 3x^2 = 3x^2\cos (x^3).$

(problem 14) Compute $\frac {d}{dx} \sin (3x^5) = \answer {\cos (3x^5)\cdot 15x^4}.$

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, $g'(x)$

example 15 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 $g'(x)$ . We have, $g'(x) =\cos (x); \quad \text {next}$ $f'(x) = 3x^2, \quad \text {so}$ $f'(g(x)) = 3g^2(x) = 3\sin ^2(x).$ By the chain rule, $h'(x) = f'(g(x))g'(x) = 3\sin ^2(x)\cos (x).$

Here is a video of example 15:

(problem 15) Compute $\frac {d}{dx} \sin ^4(x) = \answer {4\sin ^3(x)\cos (x)}.$

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, $g'(x)$

example 16 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 $g'(x)$ . We have, $g'(x) = -2x; \quad \text {next}$ $f'(x) = e^x, \quad \text {so}$ $f'(g(x)) = e^{g(x)} = e^{-x^2}.$ 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 example 16:

(problem 16)

Compute $\frac {d}{dx} e^{\sqrt x} = \answer {\frac {e^{\sqrt x}}{2\sqrt x}}$

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, $g'(x)$

example 17 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 $g'(x)$ . We have, $g'(x) = 4x^3; \quad \text {next}$ $f'(x) = \frac {1}{x}, \quad \text {so}$ $f'(g(x)) =\frac {1}{g(x)} = \frac {1}{x^4 + 1} .$ 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 example 17:

(problem 17) Compute $\frac {d}{dx} \ln (x^3 -2) = \answer {\frac {3x^2}{x^3 - 2}}.$

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, $g'(x)$

example 18 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 $g'(x)$ . We have, $g'(x) = \frac {1}{x}; \quad \text {next}$ $f'(x) = 5x^4, \quad \text {so}$ $f'(g(x)) = 5g^4(x) = 5\ln ^4(x).$ 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 example 18:

(problem 19)

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

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, $g'(x)$

example 20 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 $g'(x)$ . We have, $g'(x) = \cos (x); \quad \text {next}$ $f'(x) = \frac {1}{x}, \quad \text {so}$ $f'(g(x)) = \frac {1}{g(x)} = \frac {1}{\sin (x)}.$ 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 example 20:

(problem 20a) Compute $\frac {d}{dx} \ln (\sec (x)) = \answer {\tan (x)}$

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, $g'(x)$

(problem 20b) Compute $\frac {d}{dx} \cos (\ln (x)) = \answer {\frac {-\sin (\ln (x))}{x}}.$

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, $g'(x)$

We now create a general chain rule formula for the natural logarithm function. With $g(x)$ as the inside function, and we obtain the following formula for the derivative of $\ln (g(x))$ : $\frac {d}{dx} \left [\ln (g(x))\right ] = \frac {1}{g(x)}\cdot g'(x) = \frac {g'(x)}{g(x)}.$

The above formula is both useful and important. We will see it again in the Logarithmic Differentiation section.

example 21 Find $\frac {d}{dx} \left [\ln (x^3 + 2)\right ]$ .

We use the above formula with $g(x) = x^3 + 2$ to get $\frac {d}{dx} \left [\ln (x^3 + 2)\right ] = \frac {3x^2}{x^3 + 2}.$

Here is a video of example 21

(problem 21) Compute $\frac {d}{dx} \left [\ln (x^4 + 3x - 1)\right ] = \answer {\frac {4x^3 + 3}{x^4 + 3x -1}}.$

5 Chain Rule Formulas

(problem 22) What is the derivative of $g^n(x)$ ?

$ng^{n-1}(x)$ $ng'(x)^{n-1}$ $g^n(x) g'(x)$ $ng^{n-1}(x)g'(x)$

(problem 23) What is the derivative of $e^{g(x)}$ ?

$g(x)e^{g(x)-1}$ $e^{g'(x)}$ $e^{g(x)}g'(x)$ $e^{g(x)}$

(problem 24) What is the derivative of $\ln (g(x))$ ?

$\frac {g'(x)}{x}$ $\frac {1}{g(x)}$ $\ln (g'(x))$ $\frac {g'(x)}{g(x)}$

(problem 25) What is the derivative of $\sin (g(x))$ ?

$\cos (g(x))$ $\sin (x) g'(x)$ $\sin (g(x))\cos (x)$ $\cos (g(x))g'(x)$

(problem 26) What is the derivative of $\cos (g(x))$ ?

$\sin (g(x))$ $\sin (g(x))g'(x)$ $\cos (x) g'(x)$ $-\sin (g(x))g'(x)$

(problem 27) What is the derivative of $\tan (g(x))$ ?

$\sec ^2(g(x))$ $\sec ^2(x) g'(x)$ $\sec ^2(g(x))g'(x)$ $\sec ^2(g'(x))$

(problem 28) What is the derivative of $\sec (g(x))$ ?

$\sec (g(x))\tan (g(x))$ $\sec (g(x))\tan (g(x))g'(x)$ $\sec (x)\tan (x)g'(x)$ $\sec (g(x))\tan (g(x))g'(x)$

(problem 29) What is the derivative of $\csc (g(x))$ ? Simplify your answer as much as possible.

$\csc (g(x))\cot (g(x))$ $\csc (g(x))\cot (g(x))g'(x)$ $-\csc (x)\cot (x)g'(x)$ $-\csc (g(x))\cot (g(x))g'(x)$

6 Proof of the Chain Rule

The chain rule is a consequence of the limit definition of the derivative. We give a proof of the chain rule here.

Chain Rule If $f$ and $g$ are differentiable functions, then the composition $f(g(x))$ is differentiable and $\frac {d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x).$

Proof: We start with the limit definition of the derivative: $\frac {d}{dx} f(g(x)) = \lim _{h \to 0} \frac {f(g(x+h)) - f(g(x))}{h}$ We multiply and divide by $g(x+h) - g(x)$ to get $\frac {d}{dx} f(g(x)) = \lim _{h \to 0} \frac {f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \frac {g(x+h) - g(x)}{h}.$ By the limit laws, this is equal to $\lim _{h \to 0} \frac {f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \lim _{h \to 0} \frac {g(x+h) - g(x)}{h}$ The second limit is $g'(x)$ by the definition of the derivative. To find the first limit, we let $k = g(x+h) - g(x)$ . By the continuity of $g$ , we have $k \to 0$ as $h \to 0$ . Thus, the first limit is $\lim _{k \to 0} \frac {f(g(x) + k) - f(g(x))}{k} = f'(g(x)).$ Putting this all together, we have $\frac {d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x).$ Note that in this proof, we had to assume that $g(x+h) \neq g(x)$ for $h \neq 0$ close to $0$ . $\blacksquare$

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

2026-03-09 03:00:55