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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives4 Marks
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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