# Chain Rule

The Chain Rule is technically a method, but it's so important that it deserves its own section.

#### What is the Chain Rule?

Suppose that we have a function $f(g(x))$, and we want to differentiate it. How would we do it?

Remember that differentiation is the slope, which is

$\frac{\text{Change in }f(g(x))}{\text{Change in } x}=\frac{d f(g(x))}{d x}.$

This isn't something we immediately know how to solve. What we do know how to solve is

$\frac{d~ f(n)}{d n}$

What if $n$ was $g(x)$? Then our numerator would be $f(g(x))$:

$\frac{d~f(g(x))}{d~g(x)}.$

But, this isn't quite what we wanted - now the denominator doesn't match. No problem - we can multiply by the differentiation of $g(x)$:

$\frac{d~f(g(x))}{\cancel{d~g(x)}}\cdot \frac{\cancel{d~g(x)}}{d x} =\frac{d f(g(x))}{d x} .$

That's it - that's the Chain Rule.

#### How long can the chain go?

Really, it's a chain - you choose the length. Rather, the problem chooses the length.

For example,

$\frac{d~d(e(f(g(x))))}{d~e(f(g(x)))}\cdot\frac{d~e(f(g(x)))}{d~f(g(x))}\cdot\frac{d~f(g(x))}{d~g(x)}\cdot \frac{d~g(x)}{d x} =\frac{d~d(e(f(g(x))))}{d x}.$

### Usefulness

Remember during practice when you had to differentiate $\ln x$? It was a pain for me to write the solution and for you to read it.

Now, we'll differentiate the same function in just a few easy steps! (No, that doesn't mean that the rest of the steps are hard).

$y=\ln x$
$x = e^y$
$\frac{dx}{dx} = \frac{d e^y}{dx}$
$\frac{dx}{dx} = \frac{d e^y}{dy}\cdot \frac{dy}{dx}$
$1 = e^y \frac{dy}{dx}$
$\frac{1}{e^y} = \frac{dy}{dx}$

Substitute from second step:

$\boxed{\frac{1}{x} = \frac{dy}{dx}}$

# Courses

© 2022 YMath.io owned and operated by Saumya Singhal