Derivative of Composite Power Functions
Composite Power Functions and Derivatives
We have previously introduced how to calculate the derivative of a function at a point using the definition of the difference quotient. We also studied how to differentiate simple functions and composite functions. Now, let’s see how to differentiate power functions of the form:
\[D [ f ( x ) ]^{g ( x )}\]To calculate the derivative of such a function, a combination of the logarithmic rule and the derivative of exponential functions is used. The general formula for the derivative of $f ( x )^{g} ( x )$, with $f$ and $g$ differentiable, is as follows:
\[D [ f ( x ) ]^{g ( x )} = f ( x )^{g ( x )} [ g^{'} ( x ) ln f ( x ) + g ( x ) \frac{f^{'} ( x )}{f ( x )} ]\]Where:
- $f ( x )^{g ( x )}$ is the original function.
- $f^{‘} ( x )$ is the derivative of $f ( x )$.
- $ln f ( x )$ is the natural logarithm of $f ( x )$.
- $g^{‘} ( x )$ is the derivative of $g ( x )$.
Example
Let’s consider the function $y = x^{2 x}$ as an example, and calculate its derivative.
First, let’s rewrite the function by applying the logarithm to both sides:
\[ln y = ln ( x^{2 x} )\]For the properties of logarithms $log_{a} ( b^{c} ) = c \cdot log_{a} ( b )$
The equality can be rewritten as:
\[ln y = 2 x \cdot ln ( x )\]Since $ln y$ is a composite function, its derivative is
\[\frac{1}{y} \cdot y^{'}\]Let’s compute the derivative for the element on the right-hand side of the equality $2 x \cdot ln ( x )$:
\[2 \cdot ln ( x ) + 2 x \cdot \frac{1}{x}\]We obtain:
\[\frac{1}{y} \cdot y^{'} = 2 \cdot ln ( x ) + 2 x \cdot \frac{1}{x}\]The equality can be rewritten as:
\[y^{'} = y \cdot ( 2 \cdot ln ( x ) + 2 )\]Since $y = x^{2 x}$, we have:
\[y^{'} = x^{2 x} \cdot ( 2 \cdot ln ( x ) + 2 )\]Therefore, the derivative of $y = x^{2}$ is equal to:
\[x^{2 x} \cdot ( 2 \cdot ln ( x ) + 2 )\]