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Derivative of the Logarithmic Function for Arbitrary Base

Problem 162: Show that \frac{d}{dx} \log_a u(x) = \frac{1}{u(x)\ln a} \frac{d}{dx} u(x), where u(x) is a differentiable function and a > 1 is an arbitrary base.

Solution: Using the relationship between \log and \ln,

(1)   \begin{equation*} \log_a u(x) = \frac{\ln u(x)}{\ln a}.\end{equation*}

Then

(2)   \begin{align*} \frac{d}{dx} \log_a u(x) & = \frac{d}{dx} \frac{\ln u(x)}{\ln a} \\ & = \frac{1}{\ln a} \frac{d}{dx} \ln u(x) \\ & = \frac{1}{\ln a} \cdot \frac{1}{u(x)} \cdot \frac{d}{dx} u(x) \quad \text{(By The Chain Rule)}\end{align*}

That is,

(3)   \begin{equation*} \frac{d}{dx} \log_a u(x) = \frac{1}{u(x)\ln a} \frac{d}{dx} u(x).$\end{equation*}

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