If f(x)=frac{x-1}{e^x},then f^{prime}(0)+f^{p | vedclass

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(C) Given that,$f(x)=\frac{x-1}{e^x} \dots (i)$Applying the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,we differentiate $f(x)

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-$1$

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(C) Given that,$f(x)=\frac{x-1}{e^x} \dots (i)$Applying the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,we differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{e^x(1) - (x-1)e^x}{(e^x)^2} = \frac{e^x(1 - x + 1)}{e^{2x}} = \frac{2-x}{e^x} \dots (ii)$Substituting $x=0$ in Eq. $(ii)$:$f'(0) = \frac{2-0}{e^0} = 2 \dots (iii)$Now,differentiate Eq. $(ii)$ with respect to $x$:$f''(x) = \frac{e^x(-1) - (2-x)e^x}{(e^x)^2} = \frac{e^x(-1 - 2 + x)}{e^{2x}} = \frac{x-3}{e^x} \dots (iv)$Substituting $x=0$ in Eq. $(iv)$:$f''(0) = \frac{0-3}{e^0} = -3 \dots (v)$Adding Eq. $(iii)$ and Eq. $(v)$:$f'(0) + f''(0) = 2 + (-3) = -1$

If $f(x)=\frac{x-1}{e^x}$,then $f^{\prime}(0)+f^{\prime \prime}(0)=$

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