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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.