Find the derivative of the following function from first principl… - Vidyadip

Question

Find the derivative of the following function from first principle. $\frac{1}{\text{x}^2}$

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Answer

Let $\text{f}(\text{x})=\frac{1}{\text{x}^2}$. Accordingly,from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{(\text{x}+\text{h})^2}-\frac{1}{\text{x}^2}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{\text{x}^2-(\text{x}+\text{h})^2}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{\text{x}^2-\text{x}^2-\text{h}^2-2\text{hx}}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-\text{h}^2-2\text{hx}}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-\text{h}-2\text{x}}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$
$=\frac{0-2\text{x}}{\text{x}^2(\text{x}+0)^2}=\frac{-2}{\text{x}^3}$

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