Question

Differentiate \(x^{\sin x}\) with respect to \(x\), while \(x>0\).

Hint:

Logarithmic differentiation is the standard method for functions of the form \(y = [f(x)]^{g(x)}\). The key is to take the natural log of both sides to bring the exponent down, then apply the product rule.

Answer 1

Step 1: Understanding the Concept:
The function is of the form \(f(x)^{g(x)}\), where both the base and the exponent are functions of \(x\).
This type of function is best differentiated using logarithmic differentiation.
Step 2: Key Formula or Approach:
The process of logarithmic differentiation involves these steps:
1. Let \(y = f(x)^{g(x)}\).
2. Take the natural logarithm (\(\ln\)) of both sides: \(\ln y = g(x) \ln(f(x))\).
3. Differentiate both sides with respect to \(x\) using the product rule and chain rule.
4. Solve for \(\frac{dy}{dx}\) and substitute the original expression for \(y\).
The product rule states that \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\).
Step 3: Detailed Explanation:
Let the given function be \(y\).
\[ y = x^{\sin x} \] Taking the natural logarithm on both sides, we get:
\[ \ln y = \ln(x^{\sin x}) \] Using the logarithm property \(\ln(a^b) = b \ln a\), we have:
\[ \ln y = (\sin x)(\ln x) \] Now, we differentiate both sides with respect to \(x\). The left side requires the chain rule, and the right side requires the product rule.
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(\sin x \cdot \ln x) \] \[ \frac{1}{y} \frac{dy}{dx} = (\sin x) \cdot \frac{d}{dx}(\ln x) + (\ln x) \cdot \frac{d}{dx}(\sin x) \] \[ \frac{1}{y} \frac{dy}{dx} = (\sin x) \cdot \left(\frac{1}{x}\right) + (\ln x) \cdot (\cos x) \] \[ \frac{1}{y} \frac{dy}{dx} = \frac{\sin x}{x} + \cos x \ln x \] To find \(\frac{dy}{dx}\), we multiply both sides by \(y\):
\[ \frac{dy}{dx} = y \left( \frac{\sin x}{x} + \cos x \ln x \right) \] Finally, substitute \(y = x^{\sin x}\) back into the equation.
\[ \frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) \] Step 4: Final Answer:
The derivative of \(x^{\sin x}\) with respect to \(x\) is \(x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)\).