Derivatives of logarithmic functions

Differentiation, derivatives and Taylor approximations: Differentiating exponential and logarithmic functions

Derivatives of logarithmic functions

Now that we know the derivative of an exponential function we can also determine the derivative of a logarithmic function. To remember:

If $f(x)=\log_a(x)$ then $\displaystyle f'(x)=\frac{1}{x\cdot\ln(a)}$ for any base $a>0$ , $a\neq 1$ .

For math enthusiasts we present the proof of $\displaystyle\bigl(\log_a(x)\bigr)'=\frac{1}{x\cdot\ln(a)}$ .

Note that the identical function $s(x)=x$ is equal to the composition of the function $f(u)=a^u$ with $u=\log_{a}(x)$ . According to the chain rule and the formula for the derivative of a power function the following is valid:

$1=s'(x)=f'\bigl(u(x)\bigr)\cdot u'(x)= a^{u}\cdot \ln(a)\cdot u'(x)=x\cdot \ln(a)\cdot u'(x)$ So

$u'(x)=\frac{1}{x\cdot \ln(a)}$

Calculate the derivative of $f(x)=\log_{2}(6x)$

$f(x)=\log_{2}(6x)$ is the composition of $y=\log_{2}(u)$ and $u=6x$ .
Then: $\displaystyle \frac{\dd y}{\dd u}=\frac{1}{\ln(2)\cdot u}$ and $\displaystyle \frac{\dd u}{\dd x}=6$ . The chain rule yields:

$\begin{aligned}f'(x)&=\frac{\dd y}{\dd u}\cdot \frac{\dd u}{\dd x}\\[0.25cm] &=\frac{1}{\ln(2)\cdot 6x}\cdot 6\\[0.25cm] &=\frac{1}{\ln(2)}\cdot \frac{1}{x}\end{aligned}$ Note that the function $f(x)$ has the following property: $f'(x)$ is a multiple of $\displaystyle\frac{1}{x}$ .

New example

The $\mathrm{pH}$ as a measure of acidity is defined as

$\mathrm{pH}=-\log_{10}\biggl(\bigl[\mathrm{H}_3\mathrm{O}^{+}\bigr]\biggr)$ If you consider $\mathrm{pH}$ as a function of the concentration $\bigl[\mathrm{H}_3\mathrm{O}^{+}\bigr]$ , then the instantaneous change of $\mathrm{pH}$ is given by

$\mathrm{pH}'=\frac{\dd\,\mathrm{pH}}{\dd\,\bigl[\mathrm{H}_3\mathrm{O}^{+}\bigr]}=-\frac{1}{\bigl[\mathrm{H}_3\mathrm{O}^{+}\bigr]\cdot\ln(10)}\approx -\frac{0.434}{\bigl[\mathrm{H}_3\mathrm{O}^{+}\bigr]}$

A particular case is the derivative of the natural logarithm:

If $f(x)=\ln(x)$ then $\displaystyle f'(x)=\frac{1}{x}$ .