The proof of the derivative of natural logarithm \( \ln(x) \) is presented using the definition of the derivative. The derivative of a composite function of the form \( \ln(u(x)) \) is also included and several examples with their solutions are presented.
The definition of the derivative \( f' \) of a function \( f \) is given by the limit
\[ f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \]
Let \( f(x) = \ln(x) \) and write the derivative of \( \ln(x) \) as
\( f'(x) = \lim_{h \to 0} \dfrac{\ln(x+h)- \ln(x) }{h} \)
Use the formula \( \ln(a) - \ln(b) = \ln(\dfrac{a}{b}) \) to rewrite the derivative of \( \ln(x) \) as
\( f'(x) = \lim_{h \to 0} \dfrac{\ln(\dfrac{x+h}{x})}{h} = \lim_{h \to 0} \dfrac{1}{h} \ln(\dfrac{x+h}{x}) \)
Use power rule of logarithms ( \( a \ln y = \ln y^a \) ) to rewrite the above limit as
\( f'(x) = \lim_{h \to 0} \ln \left(\dfrac{x+h}{x}\right)^{\dfrac{1}{h}} = \lim_{h \to 0} \ln \left(1+\dfrac{h}{x}\right)^{\dfrac{1}{h}} \)
Let \( y = \dfrac{h}{x} \)
and note that
\( \lim_{h \to 0} y = 0\)
We now express h in terms of y
\(h = y x \)
With the above substitution, we can write
\( \lim_{h \to 0} \ln \left(1+\dfrac{h}{x}\right)^{\dfrac{1}{h}} = \lim_{y \to 0} \ln \left(1+y\right)^{\dfrac{1}{y x}} \)
Use power rule of logarithms ( \( \ln y^a = a \ln y \) ) to rewrite the above limit as
\( = \lim_{y \to 0} \dfrac{1}{x} \ln \left(1+y\right)^{\dfrac{1}{y}} \)
One of the definitions of the Euler Constant e is
\( e = \lim_{m \to 0} ( 1 + m) ^{\dfrac{1}{m}} \)
Hence the limit we are looking for is given by
\( \lim_{h \to 0} \ln \left(1+\dfrac{h}{x}\right)^{\dfrac{1}{h}} = \lim_{y \to 0} \dfrac{1}{x} \ln \left(1+y\right)^{\dfrac{1}{y}} = \dfrac{1}{x} \ln e = \dfrac{1}{x} \)
Conclusion: \[ \dfrac{d}{dx} \ln(x) = \dfrac{1}{x} \]
Example 1
Find the derivative of the composite natural exponential functions
Solution to Example 1