Differentiate the function: $$y = \tan\theta(\sin\theta + \cos\theta)$$
How I approached the question:
$$y = \tan\theta(\sin\theta + \cos\theta)$$
$$y = [\tan\theta\sin\theta] + [\tan\theta\cos\theta]$$
Using the Product Rule: $f'g + fg'$
$$y' = [\sec^2\theta\sin\theta + \tan\theta\cos\theta] + [\sec^2\theta\cos\theta + (\tan\theta)(-\sin\theta)]$$
$$y' = \sec^2\theta\sin\theta + \tan\theta\cos\theta + \sec^2\theta\cos\theta -\sin\theta\tan\theta$$
At this point, I have no idea how to simplify it any further.
Best Answer
Use directly the product rule: \begin{align} (\tan\theta(\sin\theta+\cos\theta))'&= \tan'\theta(\sin\theta+\cos\theta)+\tan\theta(\sin\theta+\cos\theta)'\\ &= \frac{1}{\cos^2\theta}(\sin\theta+\cos\theta)+\tan\theta(\cos\theta-\sin\theta)\\ &=\frac{\sin\theta}{\cos^2\theta}+\frac{1}{\cos\theta}+\sin\theta-\frac{\sin^2\theta}{\cos\theta}\\ &=\frac{\sin\theta}{\cos^2\theta}+\sin\theta+\cos\theta \end{align}