Derivative of a Function Raised to the Power of Another Function

   

Find the first derivative of $y = u^v$ showing all the steps.

Derivative of $u^v$

Note that in general that, a function of the form $y = u^v$, where $u$ and $v$ are functions, is neither a power function of the form $x^k$ nor an exponential function of the form $b^x$ > and therefore the common formulas of Differentiation may not be applied. here we suggest a method to find the first derivative a function of the form $y = u^v$ where $u$ and $v$ are functions whose derivatives exist.
Given $y = u^v$
Take the $\ln$ of both sides of the above
$\ln y = \ln (u^v)$
Use properties of logarithmic functions $\quad \ln u^v = v \ln u \;$ to the right side of the above equation and obtain
$\ln y = v \ln u$
Differentiate both sides of the above with respect to $x$, using chain rule and the product rule.
$\dfrac{dy}{dx} \dfrac{1}{y} = \dfrac{dv}{dx} \ln u + v \dfrac{du}{dx} \dfrac{1}{u}$
Multiply both sides by $y$
$\dfrac{dy}{dx} = y \left( \dfrac{dv}{dx} \ln u + v \dfrac{du}{dx} \dfrac{1}{u} \right)$
Substitute $y$ by $u^v$ to obtain the final answer
$\boxed {\dfrac{dy}{dx} = u^v \left( \dfrac{dv}{dx} \ln u + v \dfrac{du}{dx} \dfrac{1}{u} \right) }$

Exercises

Find the first derivative of

1. $y = (x+3)^{x - 2}$
2. $y = (x^2+2)^{\ln x + 1}$

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