Alexander Bass
This note is one of many taken during 2025

Textbook Derivatives

I know, it’s in the textbook (any textbook), but I never really learned the derivative of f(x)=log⁡n(x)f(x) = \log_n(x)

It’s not too complicated though using the logarithm change-of-base identity.

ddxlog⁡n(x)=ddxln⁡(x)ln⁡(n)=1xln⁡(n) \frac{d}{dx} \log_n(x) = \frac{d}{dx} \frac{\ln(x)}{\ln(n)} = \frac{1}{x\ln(n)}

The exponential form isn’t that bad either.

g(x)=nxln⁡(g(x))=xln⁡(n)ddxln⁡(g(x))=ddxxln⁡(n)1g(x)g′(x)=ln⁡(n)ddxnx=ln⁡(n)nx \begin{aligned} g(x) &= n^x\\ \ln(g(x)) &= x\ln(n)\\ \frac{d}{dx} \ln(g(x)) &= \frac{d}{dx} x\ln(n)\\ \frac{1}{g(x)}g'(x) &= \ln(n)\\ \frac{d}{dx}n^x &= \ln(n)n^x \end{aligned}

Of course, given that nxn^x and log⁡n(x)\log_n(x) are inverses, we can also find the exponential through the identity found in the previous section.

f(x)=log⁡n(x)g(x)=nxddxnx=11ln⁡(n)nx=ln⁡(n)nx \begin{aligned} f(x) &= \log_n(x)\\ g(x) &= n^x\\ \frac{d}{dx} n^x &= \frac{1}{\frac{1}{ \ln(n)n^x}} = \ln(n)n^x \end{aligned}
This note is one of many taken during 2025