Question

Differentiate from first principle:

$\left(ix\right)\frac{1}{\sqrt{3-x}}$

Answer 1

Given:

$f\left(x\right)=\frac{1}{\sqrt{3-x}}$

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:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{1}{\sqrt{3-(x+h)}}-\frac{1}{\sqrt{3-x}}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{1}{\sqrt{3-x-h}}-\frac{1}{\sqrt{3-x}}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}=\frac{(\sqrt{3-x}-\sqrt{3-x-h})}{h\sqrt{3-x}\sqrt{3-x-h}}$

Rationalizing the numerator, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}=\frac{(\sqrt{3-x}-\sqrt{3-x-h})}{h\sqrt{3-x}\sqrt{3-x-h}}\times \frac{(\sqrt{3-x}+\sqrt{3-x-h})}{(\sqrt{3-x}+\sqrt{3-x-h)}}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(3-x)-(3-x-h)}{h\sqrt{3-x}\sqrt{3-x-h}}\times \frac{1}{(\sqrt{3-x}+\sqrt{3-x-h})}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{\sqrt{3-x}\sqrt{3-x-h}}\times \frac{1}{\sqrt{3-x}+\sqrt{3-x-h}}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\left(\sqrt{3-x}\sqrt{3-x-0}\right)(\sqrt{3-x}+\sqrt{3-x-0})}$

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

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

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