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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY2 Marks
Question
Differentiate the following  $\text{e}^{\text{x}}+\text{e}^{\text{x}^{2}} +.....+\ \text{e}^{\text{x}^{5}}$ w.r.t.x.
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Answer

$\text{Let y} = \text{e}^{\text{x}}+\text{e}^{\text{x}^{2}} +\text{e}^{\text{x}^{3}}+\text{e}^{\text{x}^{4}}+\ \text{e}^{\text{x}^{5}}$
$\therefore\ \frac{\text{dy}}{\text{dx}} =\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}+\text{e}^{\text{x}^{2}} +\text{e}^{\text{x}^{3}}+\text{e}^{\text{x}^{4}}+\ \text{e}^{\text{x}^{5}})$
$ =\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}})+\frac{\text{d}}{\text{dx}}( \text{e}^{\text{x}^{2}})+\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}^{3}})+\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}^{4}})+\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}^{5}})$
$ =\text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}} ({\text{x}})+\text{e}^{\text{x}^{2}}.\frac{\text{d}}{\text{dx}} (\text{x}^{2})+\text{e}^{\text{x}^{3}}.\frac{\text{d}}{\text{dx}} ({\text{x}^{3})}+\text{e}^{\text{x}^{4}}.\frac{\text{d}}{\text{dx}}({\text{x}^{4}})+\text{e}^{\text{x}^{5}}.\frac{\text{d}}{\text{dx}} ({\text{x}^{5}})$
$= \text{e}^{\text{x}}.1+\text{e}^{\text{x}^{2}}.2\text{x} +\text{e}^{\text{x}^{3}}.3\text{x}^{2}+\text{e}^{\text{x}^{4}}.4\text{x}^{3}+\ \text{e}^{\text{x}^{5}}.5\text{x}^{4}$
$ =\text{e}^{\text{x}}+2 \text{x} \ \text{e}^{\text{x}^{2}} +3\text{x}^{2}\ \text{e}^{\text{x}^{3}}+\ 4\text{x}^{3}\ \text{e}^{\text{x}^{4}}+\ 5\text{x}^{4}\ \text{e}^{\text{x}^{5}}$

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