Question

Differentiate from first principle:

$\left(ii\right)\frac{1}{\sqrt{x}}$

Answer 1

Given: $f\left(x\right)=\frac{1}{\sqrt{x}}$

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ the above expression, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}$

Rationalising the numerator, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}\times \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x-(x+h)}{h\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-1}{\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{\sqrt{x}\sqrt{x}(\sqrt{x}+\sqrt{x})}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{x\times 2\sqrt{x}}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{2x^{\frac{3}{2}}}$

$\Rightarrow f^{\prime }\left(x\right)=-\frac{1}{2}{x}^{\frac{-3}{2}}$

Therefore, the derivative of $\frac{1}{\sqrt{x}}\text{is}-\frac{1}{2}{x}^{\frac{-3}{2}}.$