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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions from first principles:
$\text{e}^{\sqrt{2\text{x}}}$
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Answer

Let $\text{f(x)}=\text{e}^{\sqrt{2\text{x}}}$
$\Rightarrow\text{f}(\text{x}+\text{h})=\text{e}^{\sqrt{2(\text{x}+\text{h})}}$
$\therefore\frac{\text{d}}{\text{dx}}(\text{f(x)})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0 }\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}}-\text{e}^{\sqrt{2\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{e}^{\sqrt{2\text{x}}}\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}-1}{\text{h}}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\big(\text{e}^{2(\text{x}+\text{h})-\sqrt{2\text{x}}}-1\big)}{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}\Bigg)\bigg(\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\bigg)$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2}(\text{x}+\text{h})-\sqrt{2\text{x}}}{\text{h}}\ \Big[\text{Since},\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^\text{h}-1}{\text{h}}=1\Big]$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\times\frac{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}{\sqrt{2(\text{x}+\text{x})}+\sqrt{2\text{x}}}$
$[\text{Rationalizing the numerator]}$
$=\text{de}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2(\text{x}+\text{h})-2\text{x}}{\text{h}\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{x}+2\text{h}-2\text{x}}{\text{h}\big(\sqrt{2}(\text{x}+\text{h})+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{h}\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\text{e}^{\sqrt{2\text{x}}}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\big(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}\big)}$
$=\frac{\text{e}^{\sqrt{2\text{x}}}}{\sqrt{2\text{x}}}$
So,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\sqrt{2\text{x}}}\big)=\frac{\text{e}^{2\text{x}}}{\sqrt{2\text{x}}}$

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