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:
Derivative of a logarithmic function If \(f(x)=\log_a(x)\) then \(\displaystyle f'(x)=\frac{1}{x\cdot\ln(a)}\) for any base \(a>0\) , \(a\neq 1\).
Then: \(\displaystyle \frac{\dd y}{\dd u}=\frac{1}{\ln(11)\cdot u}\) and \(\displaystyle \frac{\dd u}{\dd x}=2\). The chain rule yields: \[\begin{aligned}f'(x)&=\frac{\dd y}{\dd u}\cdot \frac{\dd u}{\dd x}\\[0.25cm] &=\frac{1}{\ln(11)\cdot 2x}\cdot 2\\[0.25cm] &=\frac{1}{\ln(11)}\cdot \frac{1}{x}\end{aligned}\] Note that the function \(f(x)\) has the following property: \(f'(x)\) is a multiple of \(\displaystyle\frac{1}{x}\) .
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:
Derivative of ln If \(f(x)=\ln(x)\) then \(\displaystyle f'(x)=\frac{1}{x}\).
