Chain rule with arctan

Write $h(x) = \arctan(u)$ where $u = 4x^2$. The derivative of the outer function is $\frac{d}{du}[\arctan(u)] = \frac{1}{1+u^2}$, so applying the chain rule gives $h'(x) = \frac{1}{1+u^2}\cdot \frac{du}{dx}$.

Since $u = 4x^2$, we have $\frac{du}{dx} = 8x$. Substituting $u = 4x^2$ back in and combining, $h'(x) = \frac{8x}{1+(4x^2)^2} = \frac{8x}{1+16x^4}$.

The inner function was four x squared, called u in the solution. The outer function was arctangent of u, and the chain rule required differentiating this outer piece first using one over one plus u squared, then multiplying by the derivative of the inner piece, eight x.