Derivative of ln x from the definition

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Let $f(x)=\ln x$, with $x>0$. (a) Compute $f'(1)$ using the definition of derivative $f'(1)=\lim_{h\to0}\dfrac{f(1+h)-f(1)}{h}$. (b) Repeat the computation for $f'(c)$ at a generic point $c>0$.

Solution

(a) Since $f(1)=\ln 1=0$: $f'(1)=\lim_{h\to0}\frac{\ln(1+h)-\ln 1}{h}=\lim_{h\to0}\frac{\ln(1+h)}{h}=1,$ using the standard limit $\displaystyle\lim_{h\to0}\frac{\ln(1+h)}{h}=1$.

(b) At a generic $c>0$: $f'(c)=\lim_{h\to0}\frac{\ln(c+h)-\ln c}{h}=\lim_{h\to0}\frac1h\ln\!\frac{c+h}{c}=\lim_{h\to0}\frac1h\ln\!\left(1+\frac hc\right).$ Setting $t=\dfrac hc\to0$, we have $\dfrac1h=\dfrac1{ct}$, hence $f'(c)=\lim_{t\to0}\frac{1}{ct}\ln(1+t)=\frac1c\lim_{t\to0}\frac{\ln(1+t)}{t}=\frac1c.$ In particular $f'(1)=1$, consistent with part (a). $\boxed{\;f'(c)=\dfrac1c\;}$