Question
Differentiate the following from first principle:$-\text{x}$
✓
Answer
Let $\text{f}\text{(x)}=-\text{x.}$ Then, $\text{f}(\text{x}+\text{h})=(-\text{x}+\text{h})$$\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}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-(\text{x}+\text{h})+(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}-1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=-1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-(\text{x}+\text{h})+(\text{x})}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=\lim_\limits{\text{h}\rightarrow0}-1$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\text{f}(\text{x})\Big)=-1$
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