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}
The limit inside the logarithm using numerical approximation is equal to \(e\)
\[\lim_{h \to 0}\left(1+\frac{x}{h}\right)^{\frac{x}{h}}=e\]
Thus,
\[y' = \frac{1}{x}\log_b e\]
Because \(b\) and \(e\) are interchanged
\[ y' = \frac{1}{x}\log_e b =\frac{1}{x}\frac{1}{\ln b} = \frac{1}{x\ln b} \]
For the natural logarithm \(\ln x=\log_e x\)
\[\frac{d}{dx}\ln x = \frac{d}{dx}\log_e x=\frac{1}{x\ln e}=\frac{1}{x}\]
Example 1: Find the derivative of \[y=\log_2 2x^3\]
Solution:
\begin{align} y'&= \frac{1}{2x^3\ln 2}(6x^2) \\&=\frac{3}{x\ln 2}\end{align}
You can also apply the change of base formula and find the derivative in respect to the natural logarithmic function.
\[ y= \log_2 2x^3=\frac{\ln2x^3}{\ln2} \]
\begin{align} y' &= \frac{1}{\ln2}\left(\ln2x^3\right)' \\ &= \frac{1}{\ln2}\frac{1}{2x^3}(6x^2) \\ &= \frac{3}{x\ln2} \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.
