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

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

$\frac{1}{\sqrt{2-3 x}}$

Solution:

Let

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

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)

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

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

Putting values in (i), we get

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

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

Now rationalizing the numerator by multiplying and divide by the conjugate

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

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

Using the formula

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

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

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

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

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

Putting h = 0, we get

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

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

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

Hence,

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