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# Derivative of square root of x

The derivative f’ of square root of x defined by f(x)=√x is for all x strictly positive f’(x)=1 / 2√x

## Derivative of square root of x

The derivative $f’$ of the function $f(x)=\sqrt{x}$ is:

$$\forall x \in ]0,+\infty[ ,\quad f’(x) = \frac{1}{2\sqrt{x}}$$

## Proof/Demonstration

\begin{aligned} \frac{df}{dx}=&\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\\ =&\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ =&\lim _{h \rightarrow 0} \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}\\ =&\lim _{h \rightarrow 0} \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\\ =&\lim _{h \rightarrow 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\ =&\lim _{h \rightarrow 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2 \sqrt{x}} \end{aligned}

We conclude that:

$$\forall x \in ]0,+\infty[ ,\quad f’(x) = \frac{1}{2\sqrt{x}}$$