Question

Differentiate from first principle:

$\left(vii\right)\frac{x+2}{3x+5}$

Answer 1

Given:

$f\left(x\right)=\frac{x+2}{3x+5}$

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

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{x+h+2}{3(x+h)+5}-\frac{x+2}{3x+5}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(x+h+2)(3x+5)-(3x+3h+5)(x+2)}{h(3x+3h+5)(3x+5)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(x+2)(3x+5)+h(3x+5)-\left[\right(x+2\left)\right(3x+5)+3h(x+2\left)\right]}{h(3x+3h+5)(3x+5)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-h}{h(3x+3h+5)(3x+5)}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{(3x+5{)}^2}$

Hence, the derivative of $\frac{x+2}{3x+5}$ is $\frac{-1}{(3x+5{)}^2}$