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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY2 Marks
Question
If $\text{y}=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+\frac{\text{x}^4}{4!}+...\infty$ then write $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y.
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Answer

Here,
$\text{y}=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+\frac{\text{x}^4}{4!}+...\infty$
Thus
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1+\frac{2\text{x}}{2!}-\frac{3\text{x}^2}{3!}+\frac{4\text{x}^3}{4!}...\infty$
$=-1+\text{x}-\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}...\infty$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=1-\frac{2\text{x}}{2!}+\frac{3\text{x}^2}{3!}-\frac{4\text{x}^3}{4!}+...\infty$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=1-\frac{2\text{x}}{2!}+\frac{3\text{x}^2}{3!}-\frac{4\text{x}^3}{4!}+...\infty$
$=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+...\infty$
$=\text{y}$

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