If frac{dy}{dx} = 4 and frac{d^2y}{dx^2} = -3 | vedclass

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(D) We know that $\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}$. Differentiating both sides with respect to $y$ using the chain rule: $\frac{d^2x}{

Vedclass · Accepted Answer

$\frac{3}{64}$

Vedclass · Answer

(D) We know that $\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}$. Differentiating both sides with respect to $y$ using the chain rule: $\frac{d^2x}{dy^2} = \frac{d}{dy} \left(\left(\frac{dy}{dx}\right)^{-1}\right) = -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d}{dy} \left(\frac{dy}{dx}\right)$. Since $\frac{d}{dy} \left(\frac{dy}{dx}\right) = \frac{d}{dx} \left(\frac{dy}{dx}\right) \cdot \frac{dx}{dy} = \frac{d^2y}{dx^2} \cdot \frac{1}{\frac{dy}{dx}}$, we have: $\frac{d^2x}{dy^2} = -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d^2y}{dx^2} \cdot \left(\frac{dy}{dx}\right)^{-1} = -\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^3}$. Given $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at point $P$: $\left(\frac{d^2x}{dy^2}\right)_P = -\frac{-3}{(4)^3} = \frac{3}{64}$.

If $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at a point $P$ on the curve $y = f(x)$, then $\left(\frac{d^2x}{dy^2}\right)_P = $

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