Find the first derivative of \( y = x^x \) for \( x > 0 \) with all the steps presented.
Derivative of \(x^{x}\) with Steps
Note that the function \( y = x^{x} \) is neither a power function of the form \( x^{k} \) nor an exponential function of the form \( b^{x} \), and the known formulas of differentiation of these two functions cannot be used directly. We need to find another method to find the first derivative of the given function.
Given:
\[ y = x^{x} \]Take the natural logarithm (\( \ln \)) of both sides:
\[ \ln y = \ln\left(x^{x}\right) \]Use properties of logarithmic functions \( \ln A^{b} = b \ln A \) on the right side:
\[ \ln y = x \ln x \]Differentiate both sides with respect to \( x \), using the chain rule on the left side and the product rule on the right:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(x \ln x) \] \[ \frac{1}{y} \cdot \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} \]Simplify the right side:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1 \]Multiply both sides by \( y \) and substitute \( y = x^x \):
\[ \frac{dy}{dx} = y(\ln x + 1) \]Exercises and Answers
Practice Problems & Solutions
Find the first derivative of:
- \( f(x) = x^{2x} \)
- \( g(x) = (\sin x)^{x} \)
- \( h(x) = (x^2 + 1)^{x} \)
Answers to the Above Exercises:
- \( f'(x) = 2x^{2x}(\ln x + 1) \)
- \( g'(x) = (\sin x)^{x}\left(\ln(\sin x) + \dfrac{x \cos x}{\sin x}\right) \)
- \( h'(x) = (x^2 + 1)^{x}\left(\ln(x^2 + 1) + \dfrac{2x^2}{x^2 + 1}\right) \)