Question

1. $y = 3 \sin^2 2x$, $\frac{d^2 y}{dx^2} = $

Explanation:

Step1: Simplify the function using trigonometric identity

We know that \(\sin^{2}\theta=\frac{1 - \cos2\theta}{2}\). Let \(\theta = 2x\), then \(y = 3\sin^{2}2x=3\times\frac{1-\cos4x}{2}=\frac{3}{2}-\frac{3}{2}\cos4x\)

Step2: Find the first derivative \(\frac{dy}{dx}\)

The derivative of a constant is \(0\), and the derivative of \(\cos u\) with respect to \(x\) is \(-\sin u\times u'\) (chain rule). Here \(u = 4x\), \(u'=4\). So \(\frac{dy}{dx}=0-\frac{3}{2}\times(-\sin4x)\times4 = 6\sin4x\)

Step3: Find the second derivative \(\frac{d^{2}y}{dx^{2}}\)

The derivative of \(\sin u\) with respect to \(x\) is \(\cos u\times u'\) (chain rule). Here \(u = 4x\), \(u' = 4\). So \(\frac{d^{2}y}{dx^{2}}=6\times\cos4x\times4=24\cos4x\)

Answer:

\(24\cos4x\)