The chain rule

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Here we compute derivatives of compositions of functions

So far we have seen how to compute the derivative of a function built up from other functions by addition, subtraction, multiplication and division. There is another very important way that we combine functions: composition. The chain rule allows us to deal with this case. Consider $h(x) = \sin (1+2x).$ While there are several different ways to differentiate this function, if we let $f(x) = \sin (x)$ and $g(x) = 1+2x$ , then we can express $h(x) = f(g(x))$ . The question is, can we compute the derivative of a composition of functions using the derivatives of the constituents $f(x)$ and $g(x)$ ? To do so, we need the chain rule.

Chain Rule If $f$ and $g$ are differentiable, then $\ddx f(g(x)) = f'(g(x))g'(x).$

It will take a bit of practice to make the use of the chain rule come naturally, it is more complicated than the earlier differentiation rules we have seen. Let’s return to our motivating example.

Compute: $\ddx \sin (1+2x)$

Set $f(x) = \sin (x)$ and $g(x) = 1+2x$ , now $f'(x) = \answer [given]{\cos (x)} \qquad \text {and}\qquad g'(x) = \answer [given]{2}.$ Hence $\begin{align*} \ddx \sin(1+2x) &= \ddx f(g(x))\\ &=f'(g(x))g'(x) \\ &= \cos(\answer[given]{1+2x})\cdot \answer[given]{2}\\ &= 2\cos(1+2x). \end{align*}$

Recall that $\ddx \cos (x) = -\sin (x)$ . We showed this by using the definition of the derivative and the sum of angles formula. Now that we have the chain rule, we can verify this fact by using the chain rule.

The derivative of cosine: Take 2 $\ddx \cos (x) = -\sin (x).$

Recall that

$\cos (x) = \sin \left (\frac {\pi }{2}-x\right )$ .

Now

$\begin{align*} \ddx \cos(x) &= \ddx \sin\left(\frac{\pi}{2}-x\right)\\ \\ &=-\cos\left(\frac{\pi}{2}-x\right) \\ &= -\sin(x). \end{align*}$

When working with derivatives of trigonometric functions, we suggest you use radians for angle measure. For example, while $\sin \left ((90^\circ \right )^2) = \sin \left (\left (\frac {\pi }{2}\right )^2\right ),$ one must be careful with derivatives as $\eval {\ddx \sin \left (x^2\right )}_{x=90^\circ } \ne \underbrace {2\cdot 90\cdot \cos (90^2)}_{\text {incorrect}}$ Alternatively, one could think of $x^\circ$ as meaning $\frac {x\cdot \pi }{180}$ , as then $90^\circ = \frac {90\cdot \pi }{180} = \frac {\pi }{2}$ . In this case $2\cdot 90^\circ \cdot \cos ((90^\circ )^2) = 2\cdot \frac {\pi }{2}\cdot \cos \left (\left (\frac {\pi }{2}\right )^2\right ).$

Let’s see a more complicated chain of compositions.

Compute: $\ddx \sqrt {1+\sqrt {x}}$

Set $f(x)=\sqrt {x}$ and $g(x)=1+x$ . Hence, $\sqrt {1+\sqrt {x}}=f(g(\answer [given]{f}(x)))$ and by the chain rule we know $\ddx f(g(f(x))) = f'(g(f(x)))g'(f(x))f'(x).$ Since $f'(x) = \answer [given]{\frac {1}{2\sqrt {x}}} \qquad \text {and}\qquad g'(x) = \answer [given]{1}$ We have that $\ddx \sqrt {1+\sqrt {x}} = \frac {1}{2\sqrt {1+\sqrt {x}}}\cdot 1\cdot \answer [given]{\frac {1}{2\sqrt {x}}}.$

The chain rule allows to differentiate compositions of functions that would otherwise be difficult to get our hands on.

Compute: $\ddx \sqrt {\sin (x^2)}$

set $f(x) = \answer [given]{\sqrt {x}}$ , $g(x) = \answer [given]{\sin (x)}$ , and $h(x) = \answer [given]{x^2}$ so that $f(g(x)) = \sqrt {\sin (x^2)}$ . Now $\begin{align*} \ddx \sqrt{\sin(x^2)} &= \ddx f(g(h(x)))\\ &= f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)\\ &= \frac{1}{2}(\answer[given]{\sin(x^2)})^{-1/2} \cdot \cos(\answer[given]{x^2}) \cdot \answer[given]{2x}. \end{align*}$

Using the chain rule, the power rule, and the product rule it is possible to avoid using the quotient rule entirely.

Compute: $\ddx \frac {x^3}{x^2+1}$

Rewriting this as $\ddx x^3(x^2+1)^{-1},$ set $f(x) = \answer [given]{x^{-1}}$ and $g(x) = \answer [given]{x^2+1}$ so that $f(g(x)) = (x^2 + 1)^{-1}$ . Now $x^3(x^2+1)^{-1} = x^3 f(g(x)),$ and by the product and chain rules $\ddx x^3 f(g(x)) = \answer [given]{3x^2} \cdot f(g(x))+ \answer [given]{x^3} \cdot f'(g(x))g'(x).$ Since $f'(x) = \answer [given]{\frac {-1}{x^2}}$ and $g'(x) = \answer [given]{2x}$ , write $\ddx \frac {x^3}{x^2+1} = \frac {3x^2}{x^2+1}-\frac {2x^4}{(x^2+1)^2}.$

We can also compute derivatives with a tables of values.


x	*f(x)*	*f’(x)*	*g(x)*	*g’(x)*

$-2$	$0$	$2$	$3$	$0.75$

$0$	$0.5$	$0.25$	$3$	$1$

$3$	$3$	$-4$	$-2$	$-1$

Using the table of values above, compute $h'(3)$ when $h(x) = f(g(x))$ .

Notice that we need to find the derivative of $h(x)$ at $x=3$ , but $h(x)$ is defined terms of $f(x)$ and $g(x)$ . Since $h(x)$ is a composition of functions, we need to take the derivative of $h(x)$ using the chain rule. $\ddx h(x) = f'(g(x))g'(x)$ Now that we know what $h'(x)$ looks like, we want to evaluate at $x=3$ . $\Big [\ddx h(x) \Big ]_{x=3} = f'\Big (g(\answer [given]{3})\Big )g'(\answer [given]{3})$ Using the table of values, we can see that $\begin {aligned} f'\Big (g(3)\Big )g'(3) &= f'\Big ( \answer [given]{-2}\Big ) (\answer [given]{-1}) \\ &= \answer [given]{2} (-1)\\ &= \answer [given]{-2}. \end {aligned}$


x	*f(x)*	*f’(x)*	*g(x)*	*g’(x)*

$-2$	$0$	$2$	$3$	$0.75$

$0$	$0.5$	$0.25$	$3$	$1$

$3$	$3$	$-4$	$-2$	$-1$

Using the table above, what is $p'(x)$ at $x=-2$ when $p(x) = (g \circ f)(x)$ ?

The value of $p'(-2)$ is $\answer [given]{2}$

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