Differentiate the following from first principle: x^2+1 x - Vidyadip

Question

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

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Answer

We have, $\text{f}\text{(x)}=\frac{\text{x}^2+1}{\text{x}}$ $\because\text{f}'\text{(x)}={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{f(x+}h)-\text{f(x)}}{\text{h}}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\frac{\text{(x+h)}^2+1}{\text{(x+h)}}-\frac{\text{x}^2+1}{\text{x}}}{\text{x}}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{x}\big[\text{x}^2+\text{h}^2+2\text{xh}+1\big]-\big(\text{x}^2+1\big)\big(\text{x+h}\big)}{\text{hx}\big(\text{x+h}\big)}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{x}^3+\text{xh}^2+2\text{x}^2\text{h}+\text{x}-\text{x}^3-\text{x}-\text{x}^2\text{h}-\text{h}}{\text{h}.\text{x}\text{(x+h)}}$ $={\lim\limits_{\text{h}\rightarrow0}}\frac{\text{xh}+2\text{x}^2-\text{x}^2-1}{\text{x(x+h)}}$ $=\frac{\text{x}^2-1}{\text{x}^2}$ $=1-\frac{1}{\text{x}^2}$

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