# Proof of Derivative of ln(x)

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.

## Proof of the Derivative of $\ln(x)$ Using the Definition of the Derivative

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: $\dfrac{d}{dx} \ln(x) = \dfrac{1}{x}$

## Derivative of the Composite Function $y = \ln(u(x))$

We now consider the composite natural logarithm of another function u(x). Use the chain rule of differentiation to write
$\displaystyle \dfrac{d}{dx} \ln(u(x)) = \dfrac{d}{du} \ln(u(x)) \dfrac{d}{dx} u$
Simplify
$= \dfrac {1}{u} \dfrac{d}{dx} u$
Conclusion

$\displaystyle \dfrac{d}{dx} \ln(u(x)) = \dfrac{1}{u} \dfrac{d}{dx} u$

Example 1
Find the derivative of the composite natural exponential functions

1. $f(x) = \ln\left(\dfrac{x^2}{x-2}\right)$
2. $g(x) = \ln (\sqrt{x^3+1})$
3. $h(x) = \ln ( x^2+2x-5 )$

Solution to Example 1

1. Let $u(x) = \left(\dfrac{x^2}{x-2}\right)$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \left(\dfrac{x^2}{x-2}\right) = \dfrac{x^2-4x}{\left(x-2\right)^2}$
Apply the rule for the composite natural logarithm function found above
$\displaystyle \dfrac{d}{dx} f(x) = \dfrac{1}{u} \dfrac{d}{dx} u = \dfrac{1}{\dfrac{x^2}{x-2}} \times \dfrac{x^2-4x}{(x-2)^2}$

$= \dfrac{x-2}{x^2} \times \dfrac{x^2-4x}{\left(x-2\right)^2} = \dfrac{(x-2)x(x-4)}{x^2(x-2)^2}$
Cancel common factors from numerator and denominator
$= \dfrac{(x-4)}{x(x-2)}$

2. Let $u(x) = \sqrt{x^3+1}$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \sqrt{x^3+1} = \dfrac{3x^2}{2\sqrt{x^3+1}}$.
Apply the above rule of differentiation for the composite natural logarithm function
$\displaystyle \dfrac{d}{dx} g(x) = \dfrac{1}{u} \dfrac{d}{dx} u = \dfrac{1}{\sqrt{x^3+1}} \times \dfrac{3x^2}{2\sqrt{x^3+1}}$
$= \dfrac{3x^2}{2(x^3+1)}$

3. Let $u(x) = x^2+2x-5$ and therefore $\dfrac{d}{dx} u = 2x+2$
Apply the rule of differentiation for the composite exponential function obtained above
$\displaystyle \dfrac{d}{dx} h(x) = \dfrac{1}{u} \dfrac{d}{dx} u = \dfrac{1}{x^2+2x-5} \times ( 2x+2)$
$= \dfrac{2x+2}{x^2+2x-5}$

## More References and links

derivative
definition of the derivative
Chain Rule of Differentiation in Calculus.