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

Question

Dr. Abubacker · Accepted Answer

(d)
$y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + ...$
$\Rightarrow y = e^{- x}$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = - e^{- x}$
Again differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{dx^{2}} = - e^{- x}( - 1) = e^{- x} = y$

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

Explanation

Reviewed by Dr. Abubacker

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