Skip to Main Content
It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results.

Calculus

Logarithmic Derivative

Using the definition of the derivative, we can show what the derivative of \(y=\log_bx\) is

\begin{align} y'&=\lim_{h \to 0} \frac{\log_b(x+h)-\log_bx}{h} \\ &=\lim_{h \to 0}\frac{\log_b\frac{x+h}{h}}{h} \end{align}

We multiply \(x\) into the top and bottom to create a special limit

\begin{align} y'&=\lim_{h \to 0}\frac{1}{x}\frac{x}{h}\log_b\left(1+\frac{x}{h}\right) \\ &=\frac{1}{x}\lim_{h \to 0}\log_b\left(1+\frac{x}{h}\right)^{\frac{x}{h}} \\ &= \frac{1}{x}\log_b\left(\lim_{h \to 0}\left(1+\frac{x}{h}\right)^{\frac{x}{h}}\right) \end{align}

We can show by numerical approximation that 

\[\lim_{h \to 0}\left(1+\frac{x}{h}\right)^{\frac{x}{h}}=e\]

Thus,

\[\frac{d}{dx}\log_bx=\frac{1}{x}\log_b e\]


For the natural logarithm \(\ln x=\log_e x\)

\[\frac{d}{dx}\ln x = \frac{d}{dx}\log_e x=\frac{1}{x}\log_e e=\frac{1}{x}\]

Examples

Example 1: Find the derivative of \[y=\log_2 2x^3\]

Solution:

\begin{align} y'&= \frac{1}{2x^3}\log_2 e\left(6x^2 \right) \\&=\frac{3}{x}\log_2 e\end{align}


Example 2: Find the derivative of \[r=\ln\frac{v^2}{v+2}\]

Solution:

\begin{align} r'&=\frac{1}{\frac{v^2}{v+2}}\left(\frac{2v(v+2)-2v}{(v+2)^2} \right) \\&=\frac{v+2}{v^2}\left(\frac{2v(v+1)}{(v+2)^2} \right) \\&=\frac{2(v+1)}{v(v+2)} \end{align}


Example 3: When air friction is considered, the time \(t\) (in \(s\)) it takes a certain falling object to attain a velocity \(v\) (in m/s) is given by \(t=5\ln \frac{5}{5-0.1v}\). Find \(dt/dv\) for \(v=10.0m/s\).

Solution: Let's find the derivative of the equation first, then substitute the velocity to find the change at the given moment.

\begin{align} \frac{dt}{dv} &=5\left(\frac{5-0.1v}{5} \right)\left(-0.5(5-0.1v)^{-2} \right) \\ &=-\frac{0.5}{5-0.1v} \end{align} 

Now we substitute in \(v=10.0m/s\)

\[ \frac{dt}{dv}=-\frac{0.5}{5-0.1(10)}=0.125\]

The change in time is 0.125 s when the velocity is at 10.0 m/s.

Creative Commons License
Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.
chat loading...