The Chain Rule

What about taking the derivative of a composite function? For example, what if h(x) = (x² + 1)⁵, where our composition f(g(x)) is f(x) = x⁵ and g(x) = (x² + 1). We would like to use the power rule but this function is clearly different that some t(x) = xⁿ. We need the Chain Rule to differentiate this function.

Theorem: If y = f(u) is a differentiable function of u, and u=g(x) is a differentiable function of x, then y=f(g(x)) is a differentiable function of x and then

$$\displaystyle\frac{dy}{dx}= \displaystyle\frac{dy}{du}\displaystyle\frac{du}{dx}$$

or,

$$\displaystyle\frac{d}{dx}\left[f(g(x))\right] = f'(g(x))g'(x)$$

Example: For the above example, y = (x² + 1)⁵, we need to use the chain rule to differentiate. we want to write y as a function of u. Therefore, the obvious choice is to let u = x² + 1. Look at the results

$$y=(u)^5 \mbox{ where } u = x^2 + 1$$

Next, to differentiate find $$\frac{dy}{du}=5(u^4)$$ and $$\frac{du}{dx}=2x$$.

And then using the chain rule,

$$\begin{eqnarray*}\displaystyle\frac{dy}{dx} &=& \displaystyle\frac{dy}{du}\displ... ...{dx}\\ &=& 5(u^4)2x\\ &=& 5(x^2+1)^4 2x\\ &=& 10x(x^2+1)^4\\ \end{eqnarray*}$$

Example: Let $f(x) = \ln(x^2 + 1)$ Then in order to rewrite f(x)as a function of a single variable u let u=x² + 1. It follows that

$$f(x) = \ln(u) \mbox{ where } u=x^2 + 1$$

Recall that $$\frac{d}{dx}\ln(x)= \displaystyle\frac{1}{x}$$. Then from the chain rule'

$$\begin{eqnarray*}\displaystyle\frac{dy}{dx} &=& \displaystyle\frac{dy}{du}\displ... ...\frac{1}{x^2+1}\right) 2x\\ &=&\displaystyle\frac{2x}{x^2+1}\\ \end{eqnarray*}$$

Notice that the Chain Rule essentially starts by taking the derivative of the outside function and multiplies by the derivative of the inside function. It can also be extended to higher compositions such as

$$f(x) = \cos^2(\ln(x^3+4))$$

See if you can get to this result

$$f'(x) = -2\cos(\ln(x^3+4))\sin(\ln(x^3+4))\left(\displaystyle\frac{1}{x^3+4}\right)3x^2$$