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
- \( y = (x+3)^{x - 2} \)
- \( y = (x^2+2)^{\ln x + 1} \)
Answer to Above Exercise:
- \( \dfrac{dy}{dx} = (x+3)^{x - 2} \; \left( \ln (x+3) + \dfrac{x - 2}{x+3} \right) \)
- \( \dfrac{dy}{dx} = (x^2+2)^{\ln x + 1} \left( \dfrac{1}{x} \ln (x^2+2) + (\ln x + 1) \dfrac{2x}{x^2+2} \right) \)
