If x=f(t) and y=g(t),then the value of frac{d | vedclass

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(D) Given,$x=f(t)$ and $y=g(t)$.First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dx}{dt} = f'(t)$ and $\frac{dy}{dt} = g

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$\frac{f^{\prime}(t) g^{\prime \prime}(t) - g^{\prime}(t) f^{\prime \prime}(t)}{\{f^{\prime}(t)\}^{3}}$

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(D) Given,$x=f(t)$ and $y=g(t)$.First,we find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dx}{dt} = f'(t)$ and $\frac{dy}{dt} = g'(t)$.Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$.Now,we find the second derivative $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g'(t)}{f'(t)} \right) = \frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) \cdot \frac{dt}{dx}$.Using the quotient rule for differentiation:$\frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) = \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^2}$.Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{f'(t)}$,we have:$\frac{d^2y}{dx^2} = \left[ \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^2} \right] \cdot \frac{1}{f'(t)} = \frac{f'(t)g''(t) - g'(t)f''(t)}{\{f'(t)\}^3}$.

If $x=f(t)$ and $y=g(t)$,then the value of $\frac{d^{2} y}{d x^{2}}$ is

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