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Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsDerivatives5 Marks
Question
Differentiate the following from first principle:$\frac{1}{\text{x}^3}$
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Answer

We have,$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\text{(x+h)}^3}-{\frac{1}{\text{x}^3}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3-\text{(x+h)}^3}{\text{x}^3\text{h(x+h)}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3-\text{(x}^3+3\text{x}^2\text{h}+3{\text{xh}^2+\text{h}^3)}}{\text{x}^3\text{h}\text{(x+h)}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3-\text{x}^3-3\text{x}^2-3\text{xh}-\text{h}^2}{\text{x}^3\text{(x+h)}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-3\text{x}^2-3\text{xh}-\text{h}^2}{\text{x}^3\text{(x+h)}^3}$
$=\frac{-3\text{x}^2}{\text{x}^6}$
$=\frac{-3}{\text{x}^4}$

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