← Back to learning objective

Derivative of ln(x) from the limit definition

Let f(x)=ln⁡(x)f(x) = \ln(x). Using the limit definition of the derivative, f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, find f′(x)f'(x). You may use the fact that lim⁡u→0ln⁡(1+u)u=1\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1.

Hint

Write f(x+h)−f(x)=ln⁡(x+h)−ln⁡(x)f(x+h) - f(x) = \ln(x+h) - \ln(x) and use the log property ln⁡A−ln⁡B=ln⁡(A/B)\ln A - \ln B = \ln(A/B) to combine into a single logarithm. Then rewrite the fraction inside the log as 1+h/x1 + h/x, and manipulate the expression so it matches the form ln⁡(1+u)/u\ln(1+u)/u with u=h/xu = h/x.

Step-by-step solution

Start with the difference quotient: f(x+h)−f(x)h=ln⁡(x+h)−ln⁡(x)h\dfrac{f(x+h)-f(x)}{h} = \dfrac{\ln(x+h) - \ln(x)}{h}. Using the log quotient rule, ln⁡(x+h)−ln⁡(x)=ln⁡(x+hx)=ln⁡(1+hx)\ln(x+h) - \ln(x) = \ln\left(\dfrac{x+h}{x}\right) = \ln\left(1 + \dfrac{h}{x}\right). So the difference quotient becomes 1hln⁡(1+hx)\dfrac{1}{h}\ln\left(1+\dfrac{h}{x}\right).

Let u=h/xu = h/x, so h=xuh = xu and as h→0h \to 0, u→0u \to 0. Rewrite the expression as 1xuln⁡(1+u)=1x⋅ln⁡(1+u)u\dfrac{1}{xu}\ln(1+u) = \dfrac{1}{x}\cdot\dfrac{\ln(1+u)}{u}. Taking the limit as h→0h \to 0 (equivalently u→0u \to 0) and using lim⁡u→0ln⁡(1+u)u=1\lim_{u\to 0}\frac{\ln(1+u)}{u} = 1, we get f′(x)=1x⋅1=1xf'(x) = \dfrac{1}{x}\cdot 1 = \dfrac{1}{x}.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap