Question

\(\dfrac{d}{dx}\big(2\cos \tfrac{3x}{4}\big)=\) ?

• A. \(-2\sin \tfrac{3x}{4}\)
• B. \(-\dfrac{3}{8}\sin \tfrac{3x}{4}\)
• C. \(-\dfrac{3}{4}\sin \tfrac{3x}{4}\)
• D. \(-\dfrac{3}{2}\sin \tfrac{3x}{4}\)

Hint:

Inside-function derivative multiplies the result: \(f(g(x))'\!=f'(g)\,g'\).

Answer 1

The Correct Option is D

Idea. Use the chain rule: derivative of \(\cos u\) is \(-\sin u\cdot u'\). The outer constant \(2\) stays outside.
Step 1. Let \(u=\frac{3x}{4}\Rightarrow u'=\frac{3}{4}\).
Step 2. \[ \frac{d}{dx}\big(2\cos u\big)=2(-\sin u)\,u'=-2\sin\!\left(\tfrac{3x}{4}\right)\cdot\frac{3}{4} =-\frac{3}{2}\sin\!\left(\tfrac{3x}{4}\right). \]