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

Solution:

Let

$f(x)=\frac{1}{\sqrt{6 x-5}}$

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)

$f(x)=\frac{1}{\sqrt{6 x-5}}$

$\mathrm{f}(\mathrm{x}+\mathrm{h})=\frac{1}{\sqrt{6 \mathrm{x}+6 \mathrm{~h}-5}}$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{1}{\sqrt{6 x+6 h-5}}-\frac{1}{\sqrt{6 x-5}}}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{\sqrt{6 x-5}-\sqrt{6 x+6 h-5}}{(\sqrt{6 x+6 h-5})(\sqrt{6 x-5})}}{h}$

Now rationalizing the numerator by multiplying and divide by the conjugate

of $\sqrt{6 x-5}-\sqrt{6 x+6 h-5}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{6 x-5}-\sqrt{6 x+6 h-5}}{h(\sqrt{6 x+6 h-5})(\sqrt{6 x-5})} \times \frac{\sqrt{6 x-5}+\sqrt{6 x+6 h-5}}{\sqrt{6 x-5}+\sqrt{6 x+6 h-5}}$

Using the formula:

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

$=\lim _{h \rightarrow 0} \frac{(\sqrt{6 x-5})^{2}-(\sqrt{6 x+6 h-5})^{2}}{h(\sqrt{6 x+6 h-5})(\sqrt{6 x-5})(\sqrt{6 x-5}+\sqrt{6 x+6 h-5})}$

$=\lim _{h \rightarrow 0} \frac{6 x-5-6 x-6 h+5}{h(\sqrt{6 x+6 h-5})(\sqrt{6 x-5})(\sqrt{6 x-5}+\sqrt{6 x+6 h-5})}$

$=\lim _{h \rightarrow 0} \frac{-6 h}{h(\sqrt{6 x+6 h-5})(\sqrt{6 x-5})(\sqrt{6 x-5}+\sqrt{6 x+6 h-5})}$

$=\lim _{h \rightarrow 0} \frac{-6}{(\sqrt{6 x+6 h-5})(\sqrt{6 x-5})(\sqrt{6 x-5}+\sqrt{6 x+6 h-5})}$

Putting h = 0, we get

$=\frac{-6}{(\sqrt{6 x+6(0)-5})(\sqrt{6 x-5})(\sqrt{6 x-5}+\sqrt{6 x+6(0)-5})}$

$=\frac{-6}{(\sqrt{6 x-5})^{2}(2 \sqrt{6 x-5})}$

$=\frac{-6}{2(\sqrt{6 x-5})^{3}}$

$=\frac{-3}{(\sqrt{6 x-5})^{3}}$

Hence,

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{-3}{(\sqrt{6 \mathrm{x}-5})^{2}}$