The chain rule allows us to take the derivative of compound
functions like $\sin\left(x^2\right)$ or $\bigl(\sin(x)\bigr)^2$.
If $f(x)=\sin(x)$ and $g(x)=x^2$, then
$\sin\left(x^2\right)=f\left(g(x)\right)$ and
$\left(\sin(x)\right)^2 = g\left(f(x)\right)$.
The chain rule says
| $$ \frac{d}{dx} \Bigl(f\bigl(g(x)\bigr)\Bigr) = f'\left(g(x)\right) \cdot g'(x)$$ |
So the derivative of $\sin\left(x^2\right)$ is $\cos\left(x^2\right)\cdot 2x$, or $2x \cos\left(x^2\right)$, while the derivative of $\sin^2(x)$ is $2 \sin(x) \cdot \cos(x)$.
It's often useful to let $u=g(x)$, so our rule becomes
| $$\frac{d}{dx}\left(f(u)\right) = f'(u) \frac{du}{dx}$$ |
Combining this with our derivatives of basic functions, we get:
| $$ \frac{d}{dx}\bigl( u^n\bigr)=nu^{n-1}\frac{du}{dx} $$ $$ \frac{d}{dx}\bigl( e^u\bigr)=e^u\frac{du}{dx} $$ $$ \frac{d}{dx}\bigl( \ln(u)\bigr)=\frac{1}{u}\frac{du}{dx} $$ $$ \frac{d}{dx}\bigl( \sin(u)\bigr)=\cos(u)\frac{du}{dx} $$ $$ \frac{d}{dx}\bigl( \cos(u)\bigr)=-\sin(u)\frac{du}{dx} $$ $$ \frac{d}{dx}\bigl(\tan(u)\bigr) =\sec^2(u)\frac{du}{dx}$$ $$ \frac{d}{dx}\bigl(\sec(u)\bigr) =\sec(u)\tan(u)\frac{du}{dx}$$ |
In particular, when taking the derivative of $\sin\left(x^2\right)$, just let $u=x^2$, so we get the derivative of $\sin(u)$ being $\cos(u)\cdot 2x = 2x\cos\left(x^2\right)$. When taking the derivative of $\sin^2(x)$, take $u=\sin(x)$, so the derivative of $u^2$ is $\displaystyle 2 u \frac{du}{dx} = 2 \sin(x)\cos(x)$.
In this form, the chain rule comes from taking $y=f(u)$, where
$u=g(x)$, so we have $$\frac{dy}{dx} = \frac{dy}{du}
\frac{du}{dx}.$$
Recall that $\displaystyle\frac{dy}{du}$ is another name for
$f'(u)=f'\left(g(x)\right)$, while $\displaystyle\frac{du}{dx}$ is
another name for $g'(x)$.