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Expansions to Remove Indeterminate Form

If y =1+ x + ...

Question

If $y = (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . \infty)$ show that $\frac{d y}{d x} = y .$

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Solution

We have , $y = e^{x} .$

$∴ \frac{d y}{d x} = \frac{d}{d x} (e^{x}) = e^{x} = y .$

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y = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + . . . . . . . \infty

, then show that

\frac{d y}{d x} = y

y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots,

\frac{d y}{d x} =

y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . . .

\frac{d y}{d x}

y = \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .,

\frac{d y}{d x} =

\sqrt{y + x} + \sqrt{y - x} = c

\frac{d y}{d x} = \frac{y}{x} - \sqrt{\frac{y^{2}}{x^{2}} - 1}

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