If $y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - \ldots,then\frac{d^{2}y}{dx^{2}}\$is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Given, $y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!}\ldots$
$\Rightarrow \ \ y\  = \ e^{- x}\ \ \ \  \Rightarrow \ \ \frac{dy}{dx} = - e^{- x}$
$	herefore\ \frac{d^{2}y}{dx^{2}} = e^{- x} = y$

Q. If $y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - \ldots,then\frac{d^{2}y}{dx^{2}}\$is equal to

Explanation

Reviewed by Dr. Abubacker

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