3.6 - The Chain Rule

#calc/theorem

Layers

• The first thing needed to understand the chain rule is the concept of layers. A function has more than one layer if it’s composed of more than one other function:
  • For example, $2x^2$ has only one layer.
  • $3{\sin }^3(4x)$ has 3 layers – $3x^3$, $\sin x$, and $4x$. First we do $4$ times some quantity $x$. We then take the $\sin$ of the result. Finally, we cube and multiple by $3$. It took us 3 steps to create this function, so there are 3 layers.

Definition of the chain rule

If $f$ is differentiable at the point $u=g(x)$, and $g$ is differentiable at
$x$, then the composite function $(f\circ g)(x)=f(g(x))$ is differentiable at $x$, and $(f\circ g{)}^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x)$.

• So basically, the derivative of the whole thing = derivative of outside(inside) * derivative of inside. This is why it’s called the inside-outside rule.

Derivative simplification rules

For when to simplify derivatives.

1. No negative exponents
2. No fractions within fractions
3. Combine all like terms

Basic example

If $a(x)={\tan }^2(x)$, then $f(x)$ is $x^2$ and $g(x)$ is $\tan x$. $a^{\prime }(x)=2(\tan x)\cdot {\sec }^{2}x$, which simplifies to ${\displaystyle \frac{2\sin x}{{\cos }^{3}x}}$.

Chain rule using Leibniz notation

If $y=f(u)$ and $u=g(x)$, then ${\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}}$ where $\frac{dy}{du}$ is evaluated at $u=g(x)$.

• More on Leibniz notation here.