The Chain Rule

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The Chain Rule

Two Forms of the Chain Rule
Version 1
Version 2
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A hybrid chain rule

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A hybrid chain rule

In addition the previous iterations of the chain rule, there is also a useful hybrid form, given below, where we assume $u$ is a function of $x$.

$$\frac{d}{dx}f(u) = f'(u) \frac{du}{dx}$$

Applications

This form allows us to expand the scope of many of our derivative formulas. In the following, we assume $u$ is a function of $x$.

The derivative of $x^n$ (with respect to $x$) is $n x^{n-1}$, so $\frac{d}{dx}(u^n)=n u^{n-1} \frac{du}{dx}$. (Notice we are differentiating with respect to $x$, but our variable is $u$, thus we need the chain rule.)

DO: Consider $\frac{d}{dx}((3x-2)^{5})$ What is u? Find this derivative.

The derivative of $\sin(x)$ is $\cos(x)$, so $\frac{d}{dx}\sin u=\cos(u) \frac{du}{dx}$.

DO: Consider $(\sin(7x^2))'$ What is u? Find this derivative.

The derivative of $\cos(x)$ is $-\sin(x)$. DO: What is $\frac{d}{dx}\cos(u)$?

The derivative of $e^x$ is $e^x$, so the derivative of $e^u$ is $e^u \frac{du}{dx}$.

DO: Find $\frac{d}{dx}e^{4x^7}$.

We often use this hybrid form to list derivative formulas including the chain rule:
$\frac{d}{dx}(\sin (u))=\cos u\frac{du}{dx}$

$\frac{d}{dx}(\cos (u))=-\sin u\frac{du}{dx}$

$\frac{d}{dx}(\tan (u))=\sec^2 u\frac{du}{dx}$

$\frac{d}{dx}(\sec (u))=\sec u\tan u\frac{du}{dx}$ etc.

Now we use the chain rule to find the derivative of a function we couldn't differentiate before.

$\frac{d}{dx}a^x=a^x\ln a$, where $a$ is a positive number.

Why? Since $a= e^{\ln(a)}$, we can rewrite $a^x$ as $(e^{\ln(a)})^x = e^{x \ln(a)}$. Taking $u=x \ln(a)$, we have \begin{eqnarray*} \displaystyle{\frac{d a^x}{dx}} &=& \displaystyle{\frac{d}{dx}(e^{x \ln(a)})} \cr &=& e^{x \ln(a)} \displaystyle{\frac{d}{dx}}\Big(x\ln(a)\Big)\cr & = & e^{x \ln(a)} \ln(a) \cr & = & a^x \ln(a). \end{eqnarray*} (Notice that $\ln(a)$ is a constant.)

Thus, including the chain rule, we get $\frac{d}{dx}(a^u)=a^u\ln a\frac{du}{dx}$.