Find the derivative of the following functions from first princip… - Vidyadip

Question

Find the derivative of the following functions from first principle:$(-\text{x})^{-1}$

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Answer

Let $\text{f(x)}=(-\text{x})^{-1}=\frac{1}{-\text{x}}=\frac{-1}{\text{x}}$. Accordingly, $\text{f(x+h)}=\frac{-1}{\text{(x+h)}}$ By first principle, $\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-1}{\text{x+h}}-\Big(\frac{-1}{\text{x}}\Big)\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-1}{\text{x+h}}+\frac{1}{\text{x}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-\text{x}+(\text{x+h})}{\text{x(x+h)}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-\text{x+x+h}}{\text{x(x+h)}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\text{h}}{\text{x(x+h)}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{x(x+h)}}$ $=\frac{1}{\text{x}.\text{x}}=\frac{1}{\text{x}^2}$

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