Derivative of Composite Power Functions

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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)} \left[ g’(x) \ln f(x) + g(x) \frac{f’(x)}{f(x)} \right ]

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 $y = x^{2x}$ as an example, and calculate its derivative.


First, let’s rewrite the function by applying the logarithm to both sides:

\ln y = \ln(x^{2x})

For the properties of logarithms $\log_a(b^c) = c \cdot \log_a(b)$

The equality can be rewritten as:

\ln y = 2x \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 $2x \cdot \ln(x)$:

2 \cdot \ln(x) + 2x \cdot \frac{1}{x}

We obtain:

\frac{1}{y} \cdot y’ = 2 \cdot \ln(x) + 2x \cdot \frac{1}{x}

The equality can be rewritten as:

y’ = y \cdot (2 \cdot \ln(x) + 2)

Since $y = x^{2x}$, we have:

y’ = x^{2x} \cdot (2 \cdot \ln(x) + 2)

Therefore, the derivative of $y = x^2$ is equal to:

x^{2x} \cdot (2 \cdot \ln(x) + 2)

Test yourself

-

\text{1. } \quad y = x^{2\cos(x)}

solution

-

\text{2. } \quad y = x^{\ln(x)}

solution

The proposed functions are designed to help you consolidate your understanding of composite function derivatives. Try solving them independently before checking the solutions provided.