(A) Given the series $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. We know that the Maclaurin series expansion for the exponential fun

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(A) Given the series $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. We know that the Maclaurin series expansion for the exponential function is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. Therefore,$y = e^x$. Now,differentiating both sides with respect to $x$,we get: $\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$. Since $y = e^x$,we have $\frac{dy}{dx} = y$.

Class 12 Mathematics Continuity and Differentiation Derivative at a point, Standard differentiation

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If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$,then $\frac{dy}{dx} = $

A
$y$
B
$y - 1$
C
$y + 1$
D
None of these

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Derivative at a point, Standard differentiation

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If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$,then $\frac{dy}{dx} = $

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$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to