If y = sin x + e^x, then frac{d^2x}{dy^2} = | vedclass

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(C) Given $y = \sin x + e^x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \cos x + e^x$. Therefore,$\frac{dx}{dy} = \frac{1}{\cos x +

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$\frac{\sin x - e^x}{(\cos x + e^x)^3}$

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(C) Given $y = \sin x + e^x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \cos x + e^x$. Therefore,$\frac{dx}{dy} = \frac{1}{\cos x + e^x} = (\cos x + e^x)^{-1} \dots (i)$. Now,differentiating $\frac{dx}{dy}$ with respect to $y$ using the chain rule: $\frac{d^2x}{dy^2} = \frac{d}{dy} [(\cos x + e^x)^{-1}] = \frac{d}{dx} [(\cos x + e^x)^{-1}] \cdot \frac{dx}{dy}$. $\frac{d^2x}{dy^2} = -1(\cos x + e^x)^{-2} \cdot (-\sin x + e^x) \cdot \frac{dx}{dy}$. Substituting the value of $\frac{dx}{dy}$ from $(i)$: $\frac{d^2x}{dy^2} = -(\cos x + e^x)^{-2} \cdot (-\sin x + e^x) \cdot (\cos x + e^x)^{-1}$. $\frac{d^2x}{dy^2} = \frac{\sin x - e^x}{(\cos x + e^x)^3}$.

If $y = \sin x + e^x,$ then $\frac{d^2x}{dy^2} = $

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