If y = e^{nx},then left( frac{d^2y}{dx^2} rig | vedclass

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(D) Given that,$y = e^{nx}$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = ne^{nx}$.Again differentiating with respect to $x$:$\frac{

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$-ne^{-nx}$

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(D) Given that,$y = e^{nx}$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = ne^{nx}$.Again differentiating with respect to $x$:$\frac{d^2y}{dx^2} = n^2 e^{nx} \quad \dots(1)$.Now,$y = e^{nx} \implies nx = \log_e y \implies x = \frac{1}{n} \log_e y$.Differentiating $x$ with respect to $y$:$\frac{dx}{dy} = \frac{1}{n} \cdot \frac{1}{y}$.Again differentiating with respect to $y$:$\frac{d^2x}{dy^2} = \frac{1}{n} \left( -\frac{1}{y^2} \right) = -\frac{1}{n(e^{nx})^2} = -\frac{1}{n e^{2nx}} \quad \dots(2)$.Multiplying equation $(1)$ and $(2)$:$\left( \frac{d^2y}{dx^2} \right) \left( \frac{d^2x}{dy^2} \right) = (n^2 e^{nx}) \left( -\frac{1}{n e^{2nx}} \right) = -n e^{nx - 2nx} = -n e^{-nx}$.

If $y = e^{nx}$,then $\left( \frac{d^2y}{dx^2} \right) \left( \frac{d^2x}{dy^2} \right)$ is equal to

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