Question

Differentiate from first principle:

$\left(xv\right)\frac{2x+3}{x-2}$

Answer 1

Given:

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

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ in the above expression, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{2(x+h)+3}{x+h-2}-\frac{2x+3}{x-2}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(2x+3)(x-2)+2h(x-2)-\left[\right(2x+3\left)\right(x-2)+h(2x+3\left)\right]}{h(x+h-2)(x-2)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{2(x-2)-(2x+3)}{(x+h-2)(x-2)}$

$\Rightarrow f^{\prime }\left(x\right)=-\frac{7}{(x-2{)}^2}$

Therefore, the derivative of $\frac{2x+3}{x-2}is\frac{-7}{(x-2{)}^2}.$