1.5 The Chain Rule

From what has been discussed so far it might be tempting to think that the derivative of a function like $\sin\,2x$ is simply $\cos\,2x$, since the derivative of $\sin\,x$ is $\cos\,x$. It turns out that is not correct:

\begin{align*} \ddx\,(\sin\,2x) ~&=~ \ddx\,(2\;\sin\,x\;\cos\,x) \quad\text{(by the double-angle formula for sine)}\\[4pt] &=~ 2\;\ddx\,(\sin\,x\;\cos\,x) \quad\text{(by the Constant Multiple Rule)}\\[4pt] &=~ 2\;\left(\sin\,x \cdot \ddx\,(\cos\,x) ~+~ \cos\,x \cdot \ddx\,(\sin\,x)\right) \quad\text{(by the Product Rule)}\\[4pt] &=~ 2\;\left(\sin\,x \cdot (-\sin\,x) ~+~ \cos\,x \cdot \cos\,x\right)\\[4pt] &=~ 2\;\left(\cos^2 x ~-~ \sin^2 x\right)\\[4pt] &=~ 2\;\cos\,2x \quad\text{(by the double-angle formula for cosine)} \end{align*}

So the derivative of $\sin\,2x$ is $2\,\cos\,2x$, not $\cos\,2x$.

In other words, you cannot simply replace $x$ by $2x$ in the derivative formula for $\sin\,x$. Instead, regard $\sin\,2x$ as a composition of two functions: the sine function and the $2x$ function. That is, let $f(u) = \sin\,u$, where the variable $u$ itself represents a function of another variable $x$, namely $u(x) = 2x$. So since $f$ is a function of $u$, and $u$ is a function of $x$, then $f$ is a function of $x$, namely: $f(x) = \sin\,2x$. Since $f$ is a differentiable function of $u$, and $u$ is a differentiable function of $x$, then $\dfdu$ and $\dudx$ both exist (with $\dfdu = \cos\,u$ and $\dudx = 2$), and multiplying the derivatives shows that $f$ is a differentiable function of $x$:

\begin{align*} \frac{\df}{\cancel{\du}} \cdot \frac{\cancel{\du}}{\dx} ~&=~ \dfdx \quad\text{since the infinitesimals $\du$ cancel, so}\\[4pt] (\cos\,u) \cdot 2 ~&=~ \dfdx \quad\Rightarrow\quad \dfdx ~=~ 2\;\cos\,u ~=~ 2\;\cos\,2x \end{align*}

The above argument holds in general, and is known as the Chain Rule:

Theorem 1.11

Chain Rule: If $f$ is a differentiable function of $u$, and $u$ is a differentiable function of $x$, then $f$ is a differentiable function of $x$, and its derivative with respect to $x$ is:

\[\dfdx ~=~ \dfdu \cdot \dudx\]

Notice how simple the proof is—the infinitesimals $\du$ cancel.^[1]

The Chain Rule should make sense intuitively. For example, if $\dfdu = 4$ then that means $f$ is increasing 4 times as fast as $u$, and if $\dudx = 3$ then $u$ is increasing 3 times as fast as $x$, so overall $f$ should be increasing $12 = 4 \cdot 3$ times as fast as $x$, exactly as the Chain Rule says.

Example 1.9

Find the derivative of $f(x) = \sin\,(x^2 + x + 1)$.

Solution: The idea is to make a substitution $u = x^2 + x + 1$ so that $f(x) = \sin\,u$. By the Chain Rule,

\begin{align*} \dfdx ~&=~ \dfdu \cdot \dudx\\[4pt] &=~ \ddu\,(\sin\,u) \;\cdot\; \ddx\,(x^2 + x + 1)\\[4pt] &=~ (\cos\,u) \cdot (2x + 1)\\[4pt] &=~ (2x + 1)\,\cos\,(x^2 + x + 1) \end{align*}

after replacing $u$ by its definition as a function of $x$ in the last step; the final answer for the derivative should be in terms of $x$, not $u$.

In the Chain Rule you can think of the function in question as the composition of an “outer” function $f$ and an “inner” function $u$; first take the derivative of the “outer” function then multiply by the derivative of the “inner” function. Think of the “inner” function as a box into which you can put any function of $x$, and the “outer” function being a function of that empty box.

For instance, for the function $f(x) = \sin\,(x^2 + x + 1)$ in the previous example, think of the “outer” function as $\sin\,\Box$, where $\Box = x^2 + x + 1$ is the “inner” function, so that

\begin{align*} f(x) ~&=~ \sin\,(x^2 + x + 1)\\ &=~ \sin\,\Box\\ \dfdx ~&=~ \left(\cos\,\Box\right) \;\cdot\; \ddx\,\Box\\[4pt] &=~ \left(\cos\,\boxed{x^2 + x + 1}\right) \;\cdot\; \ddx\,\boxed{x^2 + x + 1}\\[4pt] &=~ (2x + 1)\,\cos\,(x^2 + x + 1) \end{align*}

Example 1.10

Find the derivative of $f(x) = (2x^4 - 3\cos\,x)^{10}$.

Solution: Here the “outer” function is $f(\Box) = \Box^{10}$ and the “inner” function is $\Box = u = 2x^4 - 3\cos\,x$:

\[\dfdx ~=~ \dfdu \cdot \dudx ~=~ 10\,\Box^9 \;\cdot\; \ddx\,(\Box) ~=~ 10\,(2x^4 - 3\cos\,x)^9 \;(8x^3 + 3\sin\,x)\]

Recall that the composition $f \circ g$ of two functions $f$ and $g$ is defined as $(f \circ g)(x) = f(g(x))$. Using prime notation the Chain Rule can be written as:

Theorem 1.12

Chain Rule: If $g$ is a differentiable function of $x$, and $f$ is a differentiable function on the range of $g$, then $f \circ g$ is a differentiable function of $x$, and its derivative with respect to $x$ is:

\[(f \circ g)'(x) ~=~ f'(g(x)) \;\cdot\; g'(x)\]

Using the Chain Rule, the Power Rule can be extended to include exponents that are rational numbers:^[2]

Theorem 1.13

\[\ddx\,\left(x^r\right) ~=~ r\,x^{r-1} \quad\text{for any rational number $r$}\]

To prove this, let $r = m/n$, where $m$ and $n$ are integers with $n \ne 0$. Then $y = x^r = x^{m/n} = \left(x^m\right)^{1/n}$, so that $y^n = x^m$. Taking the derivative with respect to $x$ of both sides of this equation gives

\begin{align*} \ddx\,\left(y^n\right) ~&=~ \ddx\,\left(x^m\right) \quad\text{, so evaluating the left side by the Chain Rule gives}\\[4pt] n y^{n-1} \;\cdot\; \dydx ~&=~ m x^{m-1}\\[4pt] n \left(x^{m/n}\right)^{n-1} \;\cdot\; \dydx ~&=~ m x^{m-1}\\[4pt] \dydx ~&=~ \frac{m x^{m-1}}{n x^{m - (m/n)}} ~=~ \frac{m}{n}\,x^{m - 1 - (m - (m/n))} ~=~ \frac{m}{n}\,x^{(m/n) - 1} ~=~ r\,x^{r-1} \quad\checkmark \end{align*}

Example 1.11

Find the derivative of $f(x) = \sqrt{x}$.

Solution: Since $\sqrt{x} = x^{1/2}$ then by the Power Rule:

\[\dfdx ~=~ \ddx\,\left(x^{1/2}\right) ~=~ \frac{1}{2}\,x^{1/2 - 1} ~=~ \frac{1}{2}\,x^{-1/2} ~=~ \frac{1}{2\,\sqrt{x}}\]

Example 1.12

Find the derivative of $f(x) = \frac{2}{3\sqrt{x}}$.

Solution:

\[\dfdx ~=~ \ddx\,\left(\frac{2}{3}\,x^{-1/2}\right) ~=~ \frac{2}{3} \cdot \frac{-1}{2}\,x^{-3/2} ~=~ -\frac{1}{3\,x^{3/2}}\]

Exercises

For Exercises 1-18, find the derivative of the given function.

$f(x) ~=~ (1 ~-~ 5x)^4$
$f(x) ~=~ 5\,(x^3 ~+~ x ~-~ 1)^4$

2
$f(x) ~=~ \sqrt{1 ~-~ 2x}\vphantom{{(1-x^2)}^{\tfrac{3}{2}}}$
$f(x) ~=~ (1 ~-~ x^2 )^{\tfrac{3}{2}}$

2
$f(x) ~=~ \dfrac{\sqrt{x}}{x ~+~ 1}$
$f(x) ~=~ \dfrac{\sqrt{x} ~+~ 1}{\sqrt{x} ~-~ 1}$

2
$f(t) ~=~ \left(\dfrac{1 ~-~ t}{1 ~+~ t}\right)^4\vphantom{\left(\dfrac{x^2 ~+~ 1}{x ~-~ 1}\right)^6}$
$f(x) ~=~ \left(\dfrac{x^2 ~+~ 1}{x ~-~ 1}\right)^6$

2
$f(x) ~=~ \sin^2 x$
$f(x) ~=~ \cos\,\left(\sqrt{x}\right)$

2
$f(x) ~=~ 3 \tan\,(5x)$
$f(x) ~=~ A\,\cos\,(\omega x ~+~ \phi)$ ($A$, $\omega$, $\phi$ are constants)

2
$f(x) ~=~ \sec\,(x^2)\vphantom{\left(\dfrac{1}{1 - x}\right)}$
$f(x) ~=~ \sin^2 \left(\dfrac{1}{1 - x}\right) ~+~ \cos^2 \left(\dfrac{1}{1 - x}\right)$

2
$L(\beta) ~=~ \dfrac{1}{\sqrt{1 ~-~ \beta^2}}\vphantom{\left(1 ~+~ \left(\dfrac{x - l}{s}\right)^2\right)^{-1}}$
$f(x) ~=~ \dfrac{1}{\pi s}\left(1 ~+~ \left(\dfrac{x - l}{s}\right)^2\right)^{-1}$ ($s$, $l$ are constants)

2
$f(x) ~=~ \cos\,(\cos\,x)$
$f(x) ~=~ \sqrt{1 ~+~ \sqrt{x}}$

In a certain type of electronic circuit^[3] the overall gain $A_v$ is given by

\[A_v ~=~ \frac{A_o}{1 ~-~ T}\]

where the loop gain $T$ is a function of the open-loop gain $A_o$.
1. Show that
  
  \[\frac{d \negmedspace A_v}{d\!A_o} ~=~ \frac{1}{1 ~-~ T} ~-~ \frac{A_o}{(1 ~-~ T)^2} \frac{d \negmedspace (1 - T)}{d\!A_o} ~.\]
2. In the case where $T$ is directly proportional to $A_o$, use part (a) to show that
  
  \[\frac{d \negmedspace A_v}{d\!A_o} ~=~ \frac{1}{(1 ~-~ T)^2} ~.\]
  
  (Hint: First show that $\;A_o \cdot \frac{d \negmedspace (1 - T)}{d\!A_o} ~=~ -T$.)
Show that the Chain Rule can be extended to 3 functions: if $u$ is a differentiable function of $x$, $v$ is a differentiable function of $u$, and $f$ is a differentiable function of $v$, then

\[\dfdx ~=~ \dfdv \;\cdot\; \dvdu \;\cdot\; \dudx\]

so that $f$ is a differentiable function of $x$. Notice that the 3 derivatives are linked together as in a chain (hence the name of the rule). The Chain Rule can be extended to any finite number of functions by the above technique.
In an internal combustion engine, as a piston moves downward the connecting rod rotates the crank in the clockwise direction, as shown in Figure 1.5.1 below.^[4]

Figure 1.5.1

The point $A$ can only move vertically, causing the point $B$ to move around a circle of radius $a$ centered at the point $O$, which is directly below the point $A$ and does not move. As the crank rotates it makes an angle $\theta$ with the line $\overline{OA}$. Let $l = AB$ and $s = OA$ as in the picture. Assume that all lengths are measured in centimeters, and let the time variable $t$ be measured in minutes.
1. Show that $s ~=~ a \cos\,\theta ~+~ \left(l^2 ~-~ a^2 \sin^2\theta\right)^{1/2}~$ for $0 \le \theta \le \pi$.
2. The mean piston speed is $\bar{S}_p = 2LN$, where $L = 2a$ is the piston stroke, and $N$ is the rotational velocity of the crank, measured in revolutions per minute (rpm). The instantaneous piston velocity is $S_p = \dsdt$. Let $R = l/a$. Show that for $0 \le \theta \le \pi$,
  
  \[\ABS{\frac{S_p}{\bar{S}_p}} ~=~ \frac{\pi}{2}\,\sin\,\theta \left[1 ~+~ \frac{\cos\,\theta}{\left(R^2 - \sin^2\theta\right)^{1/2}}\right] ~~.\]

Some textbooks give dire warnings to not think that $\du$ is an actual quantity that can be canceled. However, you can safely ignore those warnings, because $\du$ is just an infinitesimal and hence can be canceled! ↩
It will be shown in Chapter 2 how to define any real number as an exponent. The Power Rule extends to that case as well. ↩
This is an example of a current-differencing negative feedback amplifier. See pp.473-479 in Schilling, D.L. and C. Belove, Electronic Circuits: Discrete and Integrated, 2nd ed., New York: McGraw-Hill, Inc., 1979. ↩
See pp.43-45 in Heywood, J.B., Internal Combustion Engine Fundamentals, New York: McGraw-Hill Inc., 1988. ↩