Question

Differentiate from first principle:

$\left(vi\right)\frac{x+1}{x+2}$

Answer 1

Given:

$f\left(x\right)=\frac{x+1}{x+2}$

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

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{x+h+1}{x+h+2}-\frac{x+1}{x+2}}{h}$


$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(x+h+1)(x+2)-(x+h+2)(x+1)}{h(x+h+2)(x+2)}$


$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(x+1)(x+2)+h(x+2)-\left[\right(x+2\left)\right(x+1)+h(x+1\left)\right]}{h(x+h+2)(x+2)}$


$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{h\left[\right(x+2)-(x+1\left)\right]}{h(x+h+2)(x+2)}$

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

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{(x+h+2)(x+2)}$

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

Therefore, the derivative of $\frac{x+1}{x+2}\text{is}\frac{1}{(x+2{)}^2}.$