Let us consider the function $y=\sqrt{x^2+1}$. Notice that this is a composite function $y=\sqrt{u}$ and $u=x^2+1$. In general, a composite function can be written as $y=f(u)$ where $u$ is a function of $x$, $u=g(x)$. While we know how to differentiate $y=\sqrt{u}$ (i.e. finding $\frac{dy}{du}$) and $u=x^2+1$ (i.e. finding $\frac{du}{dx}$), we do not know how to differentiate $y=\sqrt{x^2+1}$ (i.e finding $\frac{dy}{dx}$). In this lecture, we would like to devise a way to differentiate a composite function. This is actually very important because the differentiable functions we stumble onto most of time are composite functions.

Let $y=f(u)$ and $u=g(x)$ and assume that both $\frac{dy}{du}$ and $\frac{du}{dx}$ exist. Now,

\begin{align*}

\frac{\Delta y}{\Delta x}&=\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x}\\

&=\frac{f(u+\Delta u)-f(u)}{\Delta u}\cdot\frac{g(\Delta x+x)-g(x)}{\Delta x}.

\end{align*}

Hence,

\begin{align*}

\frac{dy}{dx}&=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}\\

&=\lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u}\cdot\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}\ (\Delta u\to 0\ \mbox{as}\ \Delta x\to 0)\\

&=\frac{dy}{du}\cdot\frac{du}{dx}

\end{align*}

or

\begin{align*}

\frac{dy}{dx}&=\lim_{\Delta u\to 0}\frac{f(u+\Delta u)-f(u)}{\Delta u}\cdot\lim_{\Delta x\to 0}\frac{g(\Delta x+x)-g(x)}{\Delta x}\\

&=f'(u)g'(x).

\end{align*}

*Theorem*. [The Chain Rule]

Let $y=f(u)$ and $u=g(x)$. If both $\frac{dy}{du}$ and $\frac{du}{dx}$ exist, then $\frac{dy}{dx}$ exists and

\begin{align*}

\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\

&=f'(u)g'(x).

\end{align*}

*Remark*. The derivation of the chain rule shown above is not rigorously correct. The reason is that $\Delta u$ may become $0$. There is a more rigorous proof of the chain rule but we will not discuss that here.

*Remark*. Students commonly feel a difficulty with applying the chain rule when they learn it for the first time. The difficulty usually is not about understanding the chain rule itself but identifying the function $u=g(x)$. The candidate for $u$ is usually the function inside parentheses (or brackets) or the innermost function.

*Example*. We are now ready to find $\frac{dy}{dx}$ when $y=\sqrt{x^2+1}$. In this case, we don’t see parentheses or brackets but the innermost function is $x^2+1$. Let $u=x^2+1$. Then $y=\sqrt{u}$. Now,

\begin{align*}

\frac{dy}{du}&=\frac{1}{2\sqrt{u}}\\

&=\frac{1}{2\sqrt{x^2+1}},\\

\frac{du}{dx}&=2x.

\end{align*}

so, we have by the chain rule

$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{x}{\sqrt{x^2+1}}.$$

*Example*. Differentiate $y=(x^3-1)^{100}$.

*Solution*. The function inside parentheses is $x^3-1$. So, it is our candidate. Let $u=x^3-1$. Then $y=u^{100}.$

By the chain rule,

\begin{align*}

\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\

&=100u^{99}\cdot(3x^2)\\

&=300x^2(x^3-1)^{99}.

\end{align*}

*Example*. Find the derivative of each function.

1. $y=\sin 4x$.

*Solution*. The innermost function is $4x$. Let $u=4x$. Then $y=\sin u$. By the chain rule,

\begin{align*}

\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\

&=\cos u\cdot4\\

&=4\cos 4x.

\end{align*}

2. $y=\sqrt{\sin x}$.

*Solution*. The innermost function is $\sin x$. Let $u=\sin x$. Then $y=\sqrt{u}$. By the chain rule,

\begin{align*}

\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\

&=\frac{1}{2\sqrt{u}}\cdot\cos x\\

&=\frac{\cos x}{2\sqrt{\sin x}}.

\end{align*}

**Update**: For those who are interested, the rigorous proof of the Chain Rule can be found here.

robertaHow do you go from 1/2sqrt(x^2+1) to 2x in the first example? Is there a step in between there that was not shown?

Sung LeePost authorRead carefully. I didn’t go from one to another. Those are two separate things. One is $\frac{dy}{du}$ and the other is $\frac{du}{dx}$

robertaIt is very hard to understand your notes. Can you please add more details and EXPLAIN how you are going through the steps instead of just arriving at the end result. I need more than this to understand the concepts.

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