Question
Differentiate the following from first principle:
$\frac{\text{x}^2-1}{\text{x}}$

Answer

We have,
$\text{f(x)}=\frac{\text{x}^2-1}{\text{x}}$
$\therefore\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{(x+h)}^2-1}{\text{(x+h)}}-\frac{\text{x}^2-1}{\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}(\text{x}^2+\text{h}^2+2\text{xh}-1)-\text{(x+h)}(\text{x}^2-1)}{\text{x(x+h)h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{xh}+2\text{x}^2-\text{x}^2+1}{\text{x(x+h)}}$
$=\frac{\text{x}^2+1}{\text{x}^2}$
$=1+\frac{1}{\text{x}^2}$

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