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Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSDerivatives3 Marks
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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