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 $n x^{n1}$,
so $\frac{d}{dx}(u^n)=n u^{n1} \frac{du}{dx}$. (Notice we are
differentiating with respect to $x$, but our variable is $u$,
thus we need the chain rule.)
DO: Consider
$\frac{d}{dx}((3x2)^{5})$ 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 $\frac{d}{dx}(a^u)=a^u\ln
a\frac{du}{dx}$.
