# 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

1. $y = x^{x - 2}$
2. $y = (x+2)^{x^2 + 1}$

1. $\dfrac{dy}{dx} = x^{x - 3} \; (x \ln x + x - 2)$
2. $\left(2x\ln \left(x+2\right)+\dfrac{x^2+1}{x+2}\right)\left(x+2\right)^{x^2+1}$