Question

Find the derivatives of$f\left(x\right)=\sqrt{x}.$

Short answer

The derivative of$f\left(x\right)=\sqrt{x}is\frac{1}{2\sqrt{x}}.$

Step-by-step solution

Given Information 

The given expression is$f\left(x\right)=\sqrt{x}.$

Simplification  

$$\begin{gathered} f\text{'}\left(x\right)=\underset{x\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ f\text{'}\left(\sqrt{x}\right)=\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h} \\ =\underset{h\to 0}{\lim }\frac{(\sqrt{x+h}-\sqrt{x})\times (\sqrt{x+h}+\sqrt{x})}{h.(\sqrt{x+h}+\sqrt{x})} \\ =\underset{h\to 0}{\lim }\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} \\ =\underset{h\to 0}{\lim }\frac{1}{\sqrt{x+h}+\sqrt{x}} \\ =\frac{1}{2\sqrt{x}} \end{gathered}$$

Therefore, the derivate of$f\left(x\right)=\sqrt{x}is\frac{1}{2\sqrt{x}}.$