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If \( y=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+---+\infty \), then \( \frac{d y}{d x} \) is equal to |
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Answer» y = 1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+....\infty\) = ex (\(\because\) ex = 1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+....\)) \(\because\) \(\frac{dy}{dx}=\frac d{dx}e^x=e^x=y\) or Alternate method: y = 1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}....\infty\) \(\therefore\) \(\frac{dy}{dx}\) = 1 + \(\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}....\infty\) = 1 + x + \(\frac{x^2}{2!}+\frac{x^3}{3!}+...\infty\) = y |
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