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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives3 Marks
Question
Differentiate the following from the first principle$\text{e}^{\sqrt{2\text{x}}}$
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Answer

We have, $\text{f}(\text{x})=\text{e}^{\sqrt{2\text{x}}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}}-\text{e}^{\sqrt{2\text{x}}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{2\text{x}}}\big(\text{e}^{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}-1\big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{2\text{x}}}\big(\text{e}^{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}-1\big)}{\sqrt{2(\text{x}+\text{h})-\sqrt{2\text{x}}}}\times\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}$
Multiplying numerator and denominator by $\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}$
$\because\text{h}\rightarrow0\Rightarrow\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}\Rightarrow0$
and $\lim_\limits{\theta\rightarrow0}\frac{\text{e}^\theta-1}{\theta}=1$
$\therefore$$\lim_\limits{\text{h}\rightarrow0}$$​​​​\text{e}^{\sqrt{2\text{x}}}$$\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}$
Again Multiplying numerator and denominator by $\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}$
$\therefore\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{2\text{x}}}\times\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\times\frac{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}$
$=\text{e}^{\sqrt{2\text{x}}}\times\frac{1}{2\sqrt{2\text{x}}}$

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