Differentiation of x^(1/2): Find the Derivative of √x

Answer

Start from the derivative definition:

$$f'(x)=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}.$$

Multiply by the conjugate:

$$\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}.$$

The numerator becomes

$$(x+h)-x=h,$$

so, for $h\ne0$,

$$\frac{\sqrt{x+h}-\sqrt{x}}{h} =\frac{1}{\sqrt{x+h}+\sqrt{x}}.$$

Therefore, for $x>0$,

$$f'(x)=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}} =\boxed{\frac{1}{2\sqrt{x}}}.$$

The real-valued function $\sqrt{x}$ is defined for $x\ge0$. At $x=0$, the right-hand difference quotient is

$$\frac{\sqrt{h}-0}{h}=\frac{1}{\sqrt{h}},$$

which grows without bound as $h\to0^+$. Thus there is no finite derivative at $x=0$.