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Derivative of sqrt(x) via limit definition

Let f(x)=xf(x) = \sqrt{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}, compute f′(x)f'(x) for f(x)=xf(x) = \sqrt{x}. Show the algebraic steps used to evaluate the limit, and state the resulting formula for f′(x)f'(x).

Hint

Write x+h−xh\frac{\sqrt{x+h} - \sqrt{x}}{h} and multiply numerator and denominator by the conjugate x+h+x\sqrt{x+h} + \sqrt{x} to eliminate the square roots from the numerator before taking the limit.

Step-by-step solution

Start with f(x+h)−f(x)h=x+h−xh\frac{f(x+h)-f(x)}{h} = \frac{\sqrt{x+h} - \sqrt{x}}{h}. Multiply the numerator and denominator by the conjugate x+h+x\sqrt{x+h} + \sqrt{x} to get (x+h)−xh(x+h+x)=hh(x+h+x)\frac{(x+h) - x}{h(\sqrt{x+h}+\sqrt{x})} = \frac{h}{h(\sqrt{x+h}+\sqrt{x})}. Since h≠0h \neq 0 in the limit process, cancel the factor of hh to obtain 1x+h+x\frac{1}{\sqrt{x+h}+\sqrt{x}}.

Now take the limit as h→0h \to 0: since x+h→x\sqrt{x+h} \to \sqrt{x}, the expression approaches 1x+x=12x\frac{1}{\sqrt{x}+\sqrt{x}} = \frac{1}{2\sqrt{x}}. Therefore f′(x)=12xf'(x) = \dfrac{1}{2\sqrt{x}}, valid for x>0x > 0.

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