The Chain Rule

The Chain Rule computes the derivative of the composite of two functions. The composite $(f\circ g)(x)$ is just "g inside f" --- that is,

$(f\circ g)(x) = f\left(g(x)\right).$

(Note that this is not multiplication!)

Here are some examples:

$(x^3 + x^2 - 7x + 1)^{99} \hbox{ is } g(x) = x^3 + x^2 - 7x + 1 \hbox{ inside } f(x) = x^{99}.$

$\matrix{ (\hphantom{x^3 + x^2 - 7x + 1})^{99} \cr \uparrow \cr x^3 + x^2 - 7x + 1 \cr}$

$\dfrac{1}{x^2 - x - 1} \hbox{ is } g(x) = x^2 - x - 1 \hbox{ inside } f(x) = \dfrac{1}{x}.$

$\matrix{ \dfrac{1}{(\hphantom{x^2 - x - 1})} \cr \uparrow \cr x^2 - x - 1 \cr}$

$\sin (x^2) \hbox{ is } g(x) = x^2 \hbox{ inside } f(x) = \sin x.$

$\matrix{\sin & (\hphantom{x^2}) \cr & \uparrow \cr & x^2 \cr}$

Here's a more complicated example:

$\cos \dfrac{1}{x^2 - 2x + 5} \hbox{ is } h(x) = x^2 - 2x + 5 \hbox{ inside } g(x) = \dfrac{1}{x} \hbox{ inside } f(x) = \cos x.$

$\matrix{ \cos \left(\hphantom{\dfrac{1}{x^2 - 2x + 5}}\right) \cr \uparrow \cr \noalign{\smallskip} \cr \dfrac{1}{(\hphantom{x^2 - 2x + 5})} \cr \noalign{\smallskip} \cr \uparrow \cr x^2 - 2x + 5 \cr}$

One way to tell which function is "inside" and which is "outside" is to think about how you would plug numbers in. For example, take $p(x) = \sin (x^2)$ . What would you do to compute on your calculator? First, you'd square 1.7 --- . Next, you'd take the sine of that --- $\sin 1.89 \approx 0.94949$ .

The function you did first --- squaring --- is the inner function. The function you did second --- sine --- is the outer function.

Example. Suppose

$f(x) = \dfrac{1}{x} \quad\hbox{and}\quad g(x) = x^2 + 1.$

Compute $(f\circ g)(x)$ , $(g\circ f)(x)$ , and $(f\circ f)(x)$ .

$(f\circ g)(x) = f\left(g(x)\right) = f\left(x^2 + 1\right) = \dfrac{1}{x^2 + 1},$

$(g\circ f)(x) = g\left(f(x)\right) = g\left(\dfrac{1}{x}\right) = \dfrac{1}{\left(\dfrac{1}{x}\right)^2 + 1} = \dfrac{x^2}{1 + x^2},$

$f\circ f)(x) = f\left(f(x)\right) = f\left(\dfrac{1}{x}\right) = \dfrac{1}{\dfrac{1}{x}} = x.\quad\halmos$

The Chain Rule says that

$\der {} x (f\circ g)(x) = \der {} x f\left(g(x)\right) = f'\left(g(x)\right) g'(x).$

In words, you differentiate the outer function while holding the inner function fixed, then you differentiate the inner function.

Example. Compute $\displaystyle \der {} x (x^3 + x^2 - 7x + 1)^{99}$ .

$(x^3 + x^2 - 7x + 1)^{99}$ looks like this:

$\matrix{(\hphantom{x^3 + x^2 - 7x + 1})^{99} \cr \uparrow \cr x^3 + x^2 - 7x + 1 \cr}$

Differentiate the outer function $(\hbox{junk})^{99}$ , obtaining $99(\hbox{junk})^{98}$ . What is "junk"? It's . The first term in the Chain Rule is $99(x^3 + x^2 - 7x + 1)^{98}$ . (Notice that I differentiated the outer function, temporarily leaving the inner one untouched.)

Next, differentiate the inner function. The derivative of is .

Therefore,

$\der {} x (x^3 + x^2 - 7x + 1)^{99} = 99(x^3 + x^2 - 7x + 1)^{98} \cdot (3x^2 + 2x - 7).\quad\halmos$

Example. Compute $\displaystyle \der {} x \left(\dfrac{1}{x^2 - x - 1}\right)$ .

While it would be correct to use the Quotient Rule, it's unnecessary.

$\matrix{\der {} x \left(\dfrac{1}{x^2 - x - 1}\right) & = & -\dfrac{1}{(x^2 - x - 1)^2} & \cdot & (2x - 1) \cr & & \uparrow & & \uparrow \cr & & \hbox{the derivative of} & & \hbox{the derivative of} \cr \noalign{\vskip2pt} & & \dfrac{1}{\hbox{junk}} & & x^2 - x - 1 \cr}$

In general, you do not need to use the Quotient Rule to differentiate things of the form

$\dfrac{\hbox{number}}{\hbox{junk}} \quad\hbox{or}\quad \dfrac{\hbox{junk}}{\hbox{number}}.$

In the first case, use the Chain Rule as above. In the second case, divide the top by the number on the bottom.

Example. Compute $\displaystyle \der {} x \dfrac{1}{x + \sec x}$ .

$\der {} x \dfrac{1}{x + \sec x} = \der {} x (x + \sec x)^{-1} = -(x + \sec x)^{-2}\cdot (1 + \sec x \tan x).\quad\halmos$

Example. Compute $\displaystyle \der {} x \sin (x^2)$ .

Recall the derivative formula for sine:

$\der {} \theta \sin \theta = \cos \theta.$

$\matrix{ \der {} x \sin (x^2) & = & \left(\cos (x^2)\right) & \cdot & 2x \cr & & \uparrow & & \uparrow \cr & & \hbox{the derivative of} & & \hbox{the derivative of} \cr & & \sin (\hbox{junk}) & & x^2 \cr}\qquad \halmos$

Example. Compute $\displaystyle \der {} x \cos \dfrac{1}{x^2 - 2x + 5}$ .

Recall the derivative formula for cosine:

$\der {} \theta \cos \theta = -\sin \theta.$

$\matrix{ \hbox{Differentiating} & \cos (\hbox{junk}) & \hbox{gives} & -\sin \dfrac{1}{x^2 - 2x + 5} \cr \hbox{Differentiating} & \dfrac{1}{\hbox{junk}} & \hbox{gives} & -\dfrac{1}{(x^2 - 2x + 5)^2} \cr \hbox{Differentiating} & x^2 - 2x + 5 & \hbox{gives} & 2x - 2 \cr}$

Therefore,

$\der {} x \cos \dfrac{1}{x^2 - 2x + 5} = \left(-\sin \dfrac{1}{x^2 - 2x + 5}\right) \left(-\dfrac{1}{(x^2 - 2x + 5)^2}\right) \left(2x - 2\right).\quad\halmos$

Example. f and g are differentiable functions. A table of some values for these functions is shown below.

$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & & & $x = 3$ & & $x = 7$ & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & $f(x)$ & & 7 & & 14 & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & $g(x)$ & & -5 & & 0 & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & $f'(x)$ & & 6 & & 2 & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & \cr & $g'(x)$ & & 10 & & 11 & \cr height2pt & \omit & & \omit & & \omit & \cr \noalign{\hrule} }}$

Find $(g\circ f)'(3)$ .

By the Chain Rule,

$(g\circ f)'(3) = g'\left(f(3)\right)\cdot f'(3) = g'(7)\cdot f'(3) = 11\cdot 6 = 66.\quad\halmos$

Example. Compute $\displaystyle \der {} x \sin (\sin x)$ .

$\der {} x \sin (\sin x) = \left[\cos (\sin x)\right]\cdot \cos x.\quad\halmos$

Example. Notice that

$\der {} x \left[(\sin x)^2 + \sin (x^2)\right] = 2(\sin x)(\cos x) + 2x\cdot \cos (x^2).$

Do you understand the difference between $(\sin x)^2$ and $\sin (x^2)$ ? Here's a picture:

$\matrix{\left(\hphantom{\sin x}\right)^2 & \qquad & \sin \left(\hphantom{x^2}\right) \cr \uparrow & & \hphantom{\sin} \uparrow \cr \sin x & & \hphantom{\sin} x^2 \cr}$

In the first case, the outer function is the squaring function; in the second case, the outer function is the sine function.

Example. Recall that

$\der {} \theta \tan \theta = (\sec \theta)^2 \quad\hbox{and}\quad \der {} \theta \cot \theta = -(\csc \theta)^2.$

$\der {} x \tan \dfrac{1}{x} = \left(\sec \dfrac{1}{x}\right)^2\cdot \left(-\dfrac{1}{x^2}\right),$

$\der {} x \sqrt{\cot (3x + 1)} = \dfrac{1}{2} \left(\cot (3x + 1)\right)^{-1/2}\cdot \left[-\cot (3x + 1) \csc (3x + 1)\right]\cdot (3).\quad\halmos$

Example. Compute $\displaystyle \der {} x \left(1 + \left(1 + x^2\right)^2\right)^2$ .

Differentiate from the outside in:

$\der {} x \left(1 + \left(1 + x^2\right)^2\right)^2 = 2\left(1 + \left(1 + x^2\right)^2\right)\cdot 2\left(1 + x^2\right)\cdot (2x).\quad\halmos$

Example. Where does the graph of $f(x) = (x^2 - 2x + 7)^{-50}$ have a horizontal tangent?

$f'(x) = (-50)(x^2 - 2x + 7)^{-51}\cdot (2x - 2) = \dfrac{(-50)(2x - 2)}{(x^2 - 2x + 7)^{51}}.$

Set and solve for x:

$\dfrac{(-50)(2x - 2)}{(x^2 - 2x + 7)^{51}} = 0, \quad -50(2x - 2) = 0, \quad 2x = 2, \quad x = 1.\quad\halmos$

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Last updated: December 2, 2005

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