Question

The derivative of $f(x) = x^{\sin x}$ is}

• A. $\cos x \log x + (\sin x)/x$
• B. $x[\cos x \log x + (\sin x)/x]$
• C. $x^{\cos x} [\cos x \log x]$
• D. $x^{\sin x} [\cos x \log x + (\sin x)/x]$

Hint:

The general derivative formula for $y = f(x)^{g(x)}$ is:
\[ y' = f(x)^{g(x)} \left[ g'(x) \log f(x) + \frac{g(x) f'(x)}{f(x)} \right] \]

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To differentiate a function of the form $u(x)^{v(x)}$, we utilize logarithmic differentiation to simplify the variable exponent.

Step 2: Detailed Explanation:
Let:
\[ y = x^{\sin x} \]
Take the natural logarithm of both sides:
\[ \log y = \log\left(x^{\sin x}\right) \]
By logarithmic power rules:
\[ \log y = \sin x \cdot \log x \]
Differentiate both sides with respect to $x$ using the product rule on the right-hand side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x) \cdot \log x + \sin x \cdot \frac{d}{dx}(\log x) \]
\[ \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \log x + \sin x \cdot \frac{1}{x} \]
Multiply both sides by $y$:
\[ \frac{dy}{dx} = y \left[ \cos x \log x + \frac{\sin x}{x} \right] \]
Substitute $y = x^{\sin x}$ back:
\[ \frac{dy}{dx} = x^{\sin x} \left[ \cos x \log x + \frac{\sin x}{x} \right] \]
Therefore, the derivative is $x^{\sin x} [\cos x \log x + (\sin x)/x]$.

Step 3: Final Answer
The correct option is (D).