The *chain rule* is subtler than the previous rules, so if it *seems* trickier to you, then you're right. OK. But it is absolutely
indispensable in general and later, and already is very helpful in
dealing with polynomials.

The general assertion may be a little hard to fathom because it is of
a different nature than the previous ones. For one thing, now we will
be talking about a *composite function* instead of just adding or
multiplying functions in a more ordinary way. So, for two functions
$f$ and $g$,
$${d\over dx}\left(f(g(x)\right)=f'(g(x))\cdot g'(x)$$
There is also the standard notation
$$(f\circ g)(x)=f(g(x))$$
for this *composite function*, but using this notation doesn't accomplish so very much.

A problem in successful use of the chain rule is that often it
requires a little thought to recognize that some formula *is* (or
can be *looked at as*) a composite function. And the very nature
of the chain rule picks on weaknesses in our understanding of the
notation. For example, the function
$$F(x)=(1+x^2)^{100}$$
is really obtained by *first* using $x$ as input to the function
which *squares and adds $1$* to its input. Then the result of that
is used as input to the function which *takes the $100$th
power*. It is necessary to think about it this way or we'll make a
mistake. The derivative is evaluated as
$${d\over dx}(1+x^2)^{100}=100(1+x^2)^{99}\cdot 2x$$

To see that this is a special case of the general formula, we need to
see what corresponds to the $f$ and $g$ in the general
formula. Specifically, let
$$f(\hbox{input})=(\hbox{input})^{100}$$
$$g(\hbox{input})=1+(\hbox{input})^2$$
The reason for writing ‘input’ and not ‘x’ for the moment is to avoid
a certain kind of mistake. But we can compute that
$$f'(\hbox{input})=100(\hbox{input})^{99}$$
$$g'(\hbox{input})=2(\hbox{input})$$
The hazard here is that *the input to $f$ is not $x$, but rather
is $g(x)$*. So the general formula gives
$${d\over dx}(1+x^2)^{100}=f'(g(x))\cdot g'(x)=100g(x)^{99}\cdot 2x=
100(1+x^2)^{99}\cdot 2x$$

More examples: $${d\over dx}\sqrt{3x+2}={d\over dx}(3x+2)^{1/2}= {1\over 2}(3x+2)^{-1/2}\cdot 3$$ $${d\over dx}(3x^5-x+14)^{11}=11(3x^5-x+14)^{10}\cdot(15x^4-1)$$

It is very important to recognize situations like $${d\over dx}(ax+b)^n=n(ax+b)^{n-1}\cdot a$$ for any constants $a,b,n$. And, of course, this includes $${d\over dx}\sqrt{ax+b}={1\over 2}(ax+b)^{-1/2}\cdot a$$ $${d\over dx}{1\over ax+b}=-(ax+b)^{-2}\cdot a={-a\over (ax+b)^2}$$

Of course, this idea can be combined with polynomials, quotients, and
products to give enormous and excruciating things where we need to use
the chain rule, the quotient rule, the product rule, etc., and
possibly several times each. But this is not *hard*, merely *tedious*, since the only things we really do come in small steps. For
example:
$${d\over dx}\left({1+\sqrt{x+2} \over (1+7x)^{33}}\right)=
{(1+\sqrt{x+2})'\cdot (1+7x)^{33}
-(1+\sqrt{x+2})\cdot((1+7x)^{33})' \over ((1+7x)^{33})^2}$$
by the quotient rule, which is then
$${({1\over 2}(x+2)^{-1/2})\cdot (1+7x)^{33}
-(1+\sqrt{x+2})\cdot((1+7x)^{33})' \over ((1+7x)^{33})^2}$$
because our observations just above (*chain rule!*) tell us that
$${d\over dx}\sqrt{x+2}={1\over 2}(x+2)^{-1/2}\cdot
(x+2)' ={1\over 2}(x+2)^{-1/2}$$
Then we use the chain rule *again* to take the derivative of that
big power of $1+7x$, so the whole thing becomes
$${({1\over 2}(x+2)^{-1/2})\cdot (1+7x)^{33}
-(1+\sqrt{x+2})\cdot(33(1+7x)^{32}\cdot 7) \over ((1+7x)^{33})^2}$$

Although we *could* simplify a bit here, let's not. The point
about having to do several things in a row to take a derivative is
pretty clear without doing algebra just now.

#### Exercises

- Find ${ d \over dx }((1-x^2)^{100})$
- Find ${ d \over dx }\sqrt{x-3}$
- Find ${ d \over dx }(x^2-\sqrt{x^2-3})$
- Find ${ d \over dx }(\sqrt{x^2+x+1})$
- Find ${ d \over dx }({\root 3 \of {x^3+x^2+x+1}})$
- Find ${ d \over dx }((x^3 + \sqrt{x+1})^{10})$