Chain Rule — Question 6

Question 6 - Solution

We are given: $$f(x) = \ln\left( \sqrt{1 + \sin^2(5x)} \right)$$

Let’s simplify before differentiating: $$f(x) = \ln\left( (1 + \sin^2(5x))^{1/2} \right) = \frac{1}{2} \ln(1 + \sin^2(5x))$$

Now we differentiate: $$f'(x) = \frac{1}{2} \cdot \frac{1}{1 + \sin^2(5x)} \cdot \frac{d}{dx}[\sin^2(5x)]$$

We apply the chain rule to $\sin^2(5x)$: $$\frac{d}{dx}[\sin^2(5x)] = 2\sin(5x) \cdot \cos(5x) \cdot 5 = 10 \sin(5x)\cos(5x)$$

Therefore: $$f'(x) = \frac{1}{2} \cdot \frac{10 \sin(5x)\cos(5x)}{1 + \sin^2(5x)}$$

$$\boxed{ f'(x) = \frac{5 \sin(5x) \cos(5x)}{1 + \sin^2(5x)} }$$