Answer
$\sec^{2}(\theta^{2}-4\theta)\times(2\theta-4)$
Work Step by Step
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.
$=\sec^{2}(\theta^{2}-4\theta)\times(2\theta-4)$
