Find the derivation of each of the following from the first principle:

Solution:

Let

$f(x)=\frac{1}{\sqrt{2 x+3}}$

We need to find the derivative of f(x) i.e. f’(x)

We know that,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ …(i)

$f(x)=\frac{1}{\sqrt{2 x+3}}$

$\mathrm{f}(\mathrm{x}+\mathrm{h})=\frac{1}{\sqrt{2 \mathrm{x}+2 \mathrm{~h}+3}}$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{1}{\sqrt{2 x+2 h+3}}-\frac{1}{\sqrt{2 x+3}}}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{\sqrt{2 x+3}-\sqrt{2 x+2 h+3}}{(\sqrt{2 x+2 h+3})(\sqrt{2 x+3})}}{h}$

Now rationalizing the numerator by multiplying and divide by the conjugate of

$\sqrt{2 x+3}-\sqrt{2 x+2 h+3}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{2 x+3}-\sqrt{2 x+2 h+3}}{h(\sqrt{2 x+2 h+3})(\sqrt{2 x+3})} \times \frac{\sqrt{2 x+3}+\sqrt{2 x+2 h+3}}{\sqrt{2 x+3}+\sqrt{2 x+2 h+3}}$

Using the formula:

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\lim _{h \rightarrow 0} \frac{(\sqrt{2 x+3})^{2}-(\sqrt{2 x+2 h+3})^{2}}{h(\sqrt{2 x+2 h+3})(\sqrt{2 x+3})(\sqrt{2 x+3}+\sqrt{2 x+2 h+3})}$

$=\lim _{h \rightarrow 0} \frac{2 x+3-2 x-2 h-3}{h(\sqrt{2 x+2 h+3})(\sqrt{2 x+3})(\sqrt{2 x+3}+\sqrt{2 x+2 h+3})}$

$=\lim _{h \rightarrow 0} \frac{-2 h}{h(\sqrt{2 x+2 h+3})(\sqrt{2 x+3})(\sqrt{2 x+3}+\sqrt{2 x+2 h+3})}$

$=\lim _{h \rightarrow 0} \frac{-2}{(\sqrt{2 x+2 h+3})(\sqrt{2 x+3})(\sqrt{2 x+3}+\sqrt{2 x+2 h+3})}$

Putting h = 0, we get

$=\frac{-2}{(\sqrt{2 x+0+3})(\sqrt{2 x+3})(\sqrt{2 x+3}+\sqrt{2 x+0+3})}$

$=\frac{-2}{(\sqrt{2 x+3})^{2}(2 \sqrt{2 x+3})}$

$=\frac{-2}{2(\sqrt{2 x+3})^{3}}$

$=\frac{-1}{(\sqrt{2 x+3})^{3}}$

Hence,

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