Find Derivative of x^x
\( \) \( \) \( \) \( \)Find the first derivative of \( y = x^ x \) for \( x \gt 0 \) with all the steps.
Derivative of \( x^x \)
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. We need to find another method to find the first derivative of the given function.
Given \[ y = x^x \]
Take the \( \ln \) of both sides of the above
\[ \ln y = \ln (x^x) \; \text{for} \; x \gt 0 \]
Use properties of logarithmic functions \( \quad \ln A^b = b \ln A \; \) to the right side of the above equation and obtain
\[ \ln y = x \ln x \]
Differentiate both sides of the above with respect to \( x \), using chain rule on the left side and the product rule on the right.
\[ \dfrac{dy}{dx} \dfrac{1}{y} = \ln x + x \dfrac{1}{x} \]
Simplify the right side
\[ \dfrac{dy}{dx} \dfrac{1}{y} = \ln x + 1 \] ,
Multiply both sides by \( y \) and simplify
\[ \dfrac{dy}{dx} = (\ln x + 1)y \]
Substitute \( y \) by \( x^x \) to obtain the final answer
\[ \boxed {\dfrac{dy}{dx} = (\ln x + 1) x^x } \]
Exercises
Find the first derivative of
- \( y = x^{x - 2} \)
- \( y = (x+2)^{x^2 + 1} \)
Answer to Above Exercise:
- \( \dfrac{dy}{dx} = x^{x - 3} \; (x \ln x + x - 2) \)
- \( \left(2x\ln \left(x+2\right)+\dfrac{x^2+1}{x+2}\right)\left(x+2\right)^{x^2+1} \)
