Derivatives of Exponential and Logarithm Functions — Question 4

Question 4

Let the function $f(x) = e^{x \ln(\sin x)}$, for $0 < x < \pi$.

• Simplify the expression $f(x)$.

• Find the derivative $f'(x)$.

Question 4 - Solution

We are given: $$f(x) = e^{x \ln(\sin x)}, \quad 0 < x < \pi$$

(a) Simplify the expression:

Recall the identity: $$e^{x \ln(\sin x)} = (e^{\ln(\sin x)})^x = (\sin x)^x$$

So: $$f(x) = (\sin x)^x$$

(b) Differentiate $f(x) = (\sin x)^x$ using logarithmic differentiation:

Take natural log on both sides: $$\ln(f(x)) = x \ln(\sin x)$$

Differentiate both sides: $$\frac{1}{f(x)} \cdot f'(x) = \frac{d}{dx}[x \ln(\sin x)]$$

Apply product rule on the right-hand side: $$\frac{d}{dx}[x \ln(\sin x)] = \ln(\sin x) + x \cdot \frac{\cos x}{\sin x} = \ln(\sin x) + x \cot x$$

Multiply both sides by $f(x) = (\sin x)^x$: $$f'(x) = (\sin x)^x \cdot \left( \ln(\sin x) + x \cot x \right)$$

Final Answer: $$\boxed{f'(x) = (\sin x)^x \left( \ln(\sin x) + x \cot x \right)}$$