Chain Rule — Question 4

Question 4 - Solution

We are given: $$y = \sqrt{1 + \sin^2(3x)} = (1 + \sin^2(3x))^{1/2}$$

Step 1: Apply the chain rule:

Let: - Outer function: $u^{1/2}$ - Inner function: $u = 1 + \sin^2(3x)$

We compute: $$\frac{dy}{dx} = \frac{1}{2}(1 + \sin^2(3x))^{-1/2} \cdot \frac{d}{dx}[\sin^2(3x)]$$

Now differentiate $\sin^2(3x)$ using the chain rule: $$\frac{d}{dx}[\sin^2(3x)] = 2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x)$$

Step 2: Combine terms: $$\frac{dy}{dx} = \frac{1}{2}(1 + \sin^2(3x))^{-1/2} \cdot 6\sin(3x)\cos(3x)$$

$$\frac{dy}{dx} = \frac{3\sin(3x)\cos(3x)}{\sqrt{1 + \sin^2(3x)}}$$

$$\boxed{ \frac{dy}{dx} = \frac{3\sin(3x)\cos(3x)}{\sqrt{1 + \sin^2(3x)}} }$$