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

Let,

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

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)

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

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

Putting values in (i), we get

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

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

$=\lim _{h \rightarrow 0} \frac{6 x^{2}+4 x+6 x h+4 h+9 x+6-\left[6 x^{2}+6 x h+4 x+9 x+9 h+6\right]}{h((3 x+3 h+2)(3 x+2))}$

$=\lim _{h \rightarrow 0} \frac{6 x^{2}+4 x+6 x h+4 h+9 x+6-6 x^{2}-6 x h-4 x-9 x-9 h-6}{h((3 x+3 h+2)(3 x+2))}$

$=\lim _{h \rightarrow 0} \frac{-5 h}{h((3 x+3 h+2)(3 x+2))}$

$=\lim _{h \rightarrow 0} \frac{-5}{((3 x+3 h+2)(3 x+2))}$

Putting h = 0, we get

$=\frac{-5}{((3 x+3(0)+2)(3 x+2))}$

$=\frac{-5}{(3 x+2)(3 x+2)}$

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

Hence,

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