Derivative of A Constant Raised to a Power

Problem 161: Show that $\frac{d}{dx}a^{u(x)} = a^{u(x)} \ln a \cdot \frac{d}{dx}u(x)$ where $a > 0$ and $u$ is a differentiable function of $x$ .

Solution: Using the relationship between $e$ and $\ln$ ,

(1) $\begin{equation*} a^{u(x)} = e^{\ln a^{u(x)}} = e^{u(x) \ln a}. \end{equation*}$

Then

(2) $\begin{align*} \frac{d}{dx}a^{u(x)} & = \frac{d}{dx}e^{u(x) \ln a} \\ & = e^{u(x) \ln a} \cdot \frac{d}{dx} u(x) \ln a \quad \text{(By the Chain Rule)} \\ & =\ln a \cdot \underbrace{e^{u(x) \ln a}}_{= a^{u(x)}} \cdot \frac{d}{dx} u(x) \\ & = \ln a \cdot a^{u(x)} \cdot \frac{d}{dx} u(x). \end{align*}$

That is,

(3) $\begin{equation*} \frac{d}{dx}a^{u(x)} = a^{u(x)} \cdot \ln a \cdot \frac{d}{dx}u(x). \end{equation*}$

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Derivative of A Constant Raised to a Power

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