Question
Differentiate the following from first principle:
$\text{e}^{-\text{x}}$
✓
Answer
We have,
$\text{f(x)}=\text{e}^{-\text{x}}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-(\text{x+h})}-\text{e}^{-\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-\text{x}}\big(\text{e}^{-\text{h}}-1\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{e}^{-\text{x}}\big(\text{e}^{-\text{h}}-1\big)}{-\text{h}}$
$= -\text{e}^{-\text{x}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\text{e}^{\theta}-1}{\theta}=1\Big]$
$\text{f(x)}=\text{e}^{-\text{x}}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-(\text{x+h})}-\text{e}^{-\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{-\text{x}}\big(\text{e}^{-\text{h}}-1\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{e}^{-\text{x}}\big(\text{e}^{-\text{h}}-1\big)}{-\text{h}}$
$= -\text{e}^{-\text{x}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\text{e}^{\theta}-1}{\theta}=1\Big]$
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