Question
Differentiate the following functions from first principles:
$e^{3x}$.
$e^{3x}$.
✓
Answer
Let $f(x) = e^{3x}$
$\Rightarrow f(x + h) = e^{3(x + h)}$
$\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3(\text{x}+\text{h})}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{x}}\text{e}^{3\text{h}}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{3\text{x}}\left\{\frac{(\text{e}^{3\text{h}}-1)}{3\text{h}}\right\}\times3$
$=3\text{e}^{3\text{x}}\Big[\text{Since, }\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
Hence,
$\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})=3\text{e}^{3\text{x}}$
$\Rightarrow f(x + h) = e^{3(x + h)}$
$\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3(\text{x}+\text{h})}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{x}}\text{e}^{3\text{h}}-\text{e}^{3\text{x}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{3\text{x}}\left\{\frac{(\text{e}^{3\text{h}}-1)}{3\text{h}}\right\}\times3$
$=3\text{e}^{3\text{x}}\Big[\text{Since, }\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}=1\Big]$
Hence,
$\frac{\text{d}}{\text{dx}}(\text{e}^{3\text{x}})=3\text{e}^{3\text{x}}$
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