If left(frac{dy}{dx}right) = frac{1}{left(fra | vedclass

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(A) Given that, $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1} \dots (i)$Differentiating both sides with respect to $x$ using the chain rule:$\frac{

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(A) Given that, $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1} \dots (i)$Differentiating both sides with respect to $x$ using the chain rule:$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2} \cdot \frac{d}{dx}\left(\frac{dx}{dy}\right)$Using the chain rule $\frac{d}{dx} = \frac{dy}{dx} \cdot \frac{d}{dy}$:$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2} \cdot \left(\frac{d^2x}{dy^2} \cdot \frac{dy}{dx}\right)$Since $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1}$, we substitute this into the equation:$\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^2 \cdot \frac{d^2x}{dy^2} \cdot \frac{dy}{dx}$$\frac{d^2y}{dx^2} = -\left(\frac{dy}{dx}\right)^3 \cdot \frac{d^2x}{dy^2}$Rearranging the terms:$\frac{d^2x}{dy^2} \left(\frac{dy}{dx}\right)^3 + \frac{d^2y}{dx^2} = 0$Comparing this with the given expression $\frac{d^2x}{dy^2} \left(\frac{dy}{dx}\right)^3 + \frac{d^2y}{dx^2} = k$, we get $k = 0$.Now, evaluating the expression $e^{k f(x)} - k f(x)$:$e^{0 \cdot f(x)} - 0 \cdot f(x) = e^0 - 0 = 1 - 0 = 1$.

If $\left(\frac{dy}{dx}\right) = \frac{1}{\left(\frac{dx}{dy}\right)}$ and $\frac{d^2x}{dy^2}\left(\frac{dy}{dx}\right)^3 + \frac{d^2y}{dx^2} = k$, then $e^{k f(x)} - k f(x) =$

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