Question

q11 chain rule: composition of composition
if ( y=cos ^{2} 3 x ), then ( d y / d x= )
a ( -6 sin 3 x cos 3 x )
b ( -2 cos 3 x )
c ( 2 cos 3 x )
d ( 6 cos 3 x )

Explanation:

Step1: Let \( u = \cos 3x \), then \( y = u^{2} \)

Use the chain rule \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \)
First, find \( \frac{dy}{du} \):
Since \( y = u^{2} \), then \( \frac{dy}{du}=2u \)

Step2: Find \( \frac{du}{dx} \)

Since \( u=\cos 3x \), let \( v = 3x \), then \( u=\cos v \)
By the chain rule \( \frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx} \)
\( \frac{du}{dv}=-\sin v \), \( \frac{dv}{dx}=3 \)
So \( \frac{du}{dx}=- 3\sin 3x \)

Step3: Calculate \( \frac{dy}{dx} \)

\( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \)
Substitute \( u = \cos 3x \), \( \frac{dy}{du}=2u \) and \( \frac{du}{dx}=-3\sin 3x \)
\( \frac{dy}{dx}=2\cos 3x\cdot(-3\sin 3x)=-6\sin 3x\cos 3x \)

Answer:

A. -6 sin 3x cos 3x