Differentiate the following from first principle:e^ 3x - Vidyadip

Question

Differentiate the following from first principle:$\text{e}^{\text{3x}}$

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Answer

We have, $\text{f(x)}=\text{e}^{3\text{x}}$ $\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{(x+h)}}-\text{e}^{3\text{x}}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{(x)}}.\text{e}^{3\text{h}}-\text{e}^{3\text{x}}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{(x)}}(\text{e}^{3\text{h}}-1)}{\text{h}}$Multiplying Numerator and Denominator by 3
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{3\text{(x)}}\frac{(\text{e}^{3\text{h}}-1)}{3\text{h}}$ $\bigg[=\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^{3\text{h}}-1}{3}=1\bigg]$ $=3\text{e}^{3\text{x}}$

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