## Calculus (3rd Edition)

$\sec^{2}(\theta^{2}-4\theta)\times(2\theta-4)$
Using the chain rule, we have $\frac{d}{d\theta}(\tan(\theta^{2}-4\theta))$ $=\sec^{2}(\theta^{2}-4\theta)\times\frac{d}{d\theta}(\theta^{2}-4\theta)$ That is, we take the derivative of the main function$\times$derivative of the inner function. $=\sec^{2}(\theta^{2}-4\theta)\times(2\theta-4)$