Find the first derivative of \( y = u^v \) showing all the steps.
Derivative of \( u^v \)
Note that in general, 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 directly. Here we suggest a method to find the first derivative of a function of the form \( y = u^v \) where \( u \) and \( v \) are functions whose derivatives exist.
Given:
\[ y = u^v \]Take the natural logarithm (\( \ln \)) of both sides:
\[ \ln y = \ln (u^v) \]Use the property of logarithmic functions \( \ln(u^v) = v \ln u \) on the right side:
\[ \ln y = v \ln u \]Differentiate both sides with respect to \( x \), using the 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 \( y \) back with \( u^v \) to obtain the final answer formula:
Exercises and Answers
Practice Problems & Solutions
Find the first derivative of:
- \( y = (x+3)^{x - 2} \)
- \( y = (x^2+2)^{\ln x + 1} \)
Answers to Above Exercise:
- \( \dfrac{dy}{dx} = (x+3)^{x - 2} \left( \ln (x+3) + \dfrac{x - 2}{x+3} \right) \)
- \( \dfrac{dy}{dx} = (x^2+2)^{\ln x + 1} \left( \dfrac{1}{x} \ln (x^2+2) + (\ln x + 1) \dfrac{2x}{x^2+2} \right) \)