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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives3 Marks
Question
Differentiate the following from first principles: $\frac{2}{\text{x}}$
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Answer

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

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