Question
Differentiate the following from first principle:$(-\text{x})^{-1}$

Answer

Let $\text{f}(\text{x})=(-\text{x})^{-1}.$ Then, $\text{f}(\text{x}+\text{h})=\Big(-(\text{x}+\text{h})\Big)^{-1}$$\therefore\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\big(-(\text{x}+\text{h})^{-1}-(-\text{x})^{-1}\big)}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{-1}{\text{x}+\text{h}}+\frac{1}{\text{x}}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\frac{-\text{x}+\text{x}+\text{h}}{\text{x}(\text{x}+\text{h})}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}\text{x}(\text{x}+\text{h})}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\lim_\limits{\text{h}\rightarrow0}\frac{1}{\text{x}(\text{x}+\text{h})}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\frac{1}{\text{x}(\text{x}+0)}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{f}(\text{x})\big)=\frac{1}{\text{x}^2}$

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