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

Solution:

Let

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

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)

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

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

Putting values in (i), we get

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

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

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

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

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

Using the formula:

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

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

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

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

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

Putting h = 0, we get

$=\frac{-1}{(\sqrt{x+0+2})(\sqrt{x+2})(\sqrt{x+2}+\sqrt{x+0+2})}$

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

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

Hence,

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