Question
Differentiate the following from first principle:

$\frac{\text{x}+1}{\text{x}+2}$

Answer

We have,

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

$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{(x+h)+1}}{\text{(x+h)+2}}-\frac{\text{x}+1}{\text{x}+2}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{(x+2)}\text{(x+h+1)}-\text{(x+1)(x+h+2)}}{\text{(x+h+2)}\text{(x+2)h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{x}^2+2\text{x+xh+2h+2+x}\big)-\big(\text{x}^2+\text{xh+2x+x+h+2}\big)}{\text{(x+h+2)(x+2)h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{(x+h+2)(x+2)h}}$

$=\frac{1}{(\text{x}+2)^2}$

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