Question

if y=1+x+x^2/2!+x^3/3......infinity, than show that dy/dx=y
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Answer 1

Given,

$\longrightarrow y=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\,\dots$

Differentiating with respect to $x,$

$\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\,\dots\right)$

$\longrightarrow \dfrac{dy}{dx}=0+1+\dfrac{2x}{2!}+\dfrac{3x^2}{3!}+\dfrac{4x^3}{4!}+\,\dots$

$\longrightarrow \dfrac{dy}{dx}=1+\dfrac{2x}{2}+\dfrac{3x^2}{3\times2!}+\dfrac{4x^3}{4\times3!}+\,\dots$

$\longrightarrow \dfrac{dy}{dx}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\,\dots$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=y}}$

Another Method:-

It is true that,

$\longrightarrow e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\,\dots$

Then,

$\longrightarrow y=e^x$

We know derivative of $e^x$ with respect to $x$ is $e^x$ itself.

Therefore,

$\longrightarrow\dfrac{dy}{dx}=e^x$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=y}}$