3.6: Derivatives of Logarithms

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

This section introduced two new derivative rules:

$\displaystyle \frac {d}{dx}\ln (x) = \frac {1}{x}$ .
$\displaystyle \frac {d}{dx}\log _a(x) = \frac {1}{x}\cdot \frac {1}{\ln (a)}$ .

We can just add this to our toolkit of derivatives that we know. All the usual rules apply.

If $f(x) = \ln (x^2+x)$ , then $f'(x) = \answer {\frac {2x+1}{x^2+x}}$ .

Notice, in the previous problem, we could write $x^2+x = x(x+1)$ and use log rules to split up the log first: $\ln (x^2+x) = \ln (x) + \ln (\answer {x+1}).$ So $f(x) =\ln (x)+\ln (x+1).$ If we take a derivative now, we get $f'(x) = \frac {1}{x}+\frac {1}{\answer {x+1}}.$ This simplifies to the answer above.

If $f(x) = \log _2(\sin (2x)+x)$ , then $f'(x) = \answer {\frac {2\cos (2x)+1}{\ln (2)(\sin (2x)+x)}}.$

If $f(x) = \frac {x\ln (x)}{x^2+1}$ , then $f'(x) = \answer {\frac {(x^2+1)(1+\ln (x))-2x^2\ln (x)}{(x^2+1)^2}}.$

If $f(x) = x\sin (\pi x)$ , then the equation of the tangent line to $y=f(x)$ at $x=1$ has equation $y = \answer {-\pi x + \pi }.$

If $f(x) = \sqrt {\ln (2x)}$ , then $f'(x) = \answer {\frac {1}{2x\sqrt {\ln (2x)}}}.$

If $f(x) = x^x$ , find $f'(x)$ .

Notice that power rule doesn’t apply because the exponent is not a number, and exponential rules don’t apply since the base is not a number. We will use logs and implicit differentiation to help us take the derivative. Writing $y =x^x$ , we can take logs of both sides and use log rules to simplify: $\ln (y) = \ln (x^x) = x\ln (x).$ Now we can take a derivative with respect to $x$ of both sides, using implicit differentiation: $\frac {1}{y}y' = \answer {1+\ln (x)},$ so $y' = y(1+\ln (x)).$ Since $y = x^x$ , we can put this in and get $y' = \answer {x^x(1+\ln (x))}.$

If $f(x) = (\sin (x))^{\ln (x)}$ , then $f'(x) =\answer {(\sin (x))^{\ln (x)}(\frac {\ln (x)\cos (x)}{\sin (x)}+\frac {\ln (\sin (x))}{x})}.$