If x=f(t) and y=g(t) are differentiable funct | vedclass

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(A) Given $x=f(t)$ and $y=g(t)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dx}{dt} = f^{\prime}(t)$ and $\frac{dy}{dt

Vedclass · Accepted Answer

$\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) \cdot f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^3}$

Vedclass · Answer

(A) Given $x=f(t)$ and $y=g(t)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dx}{dt} = f^{\prime}(t)$ and $\frac{dy}{dt} = g^{\prime}(t)$.$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g^{\prime}(t)}{f^{\prime}(t)}$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g^{\prime}(t)}{f^{\prime}(t)} \right) = \frac{d}{dt} \left( \frac{g^{\prime}(t)}{f^{\prime}(t)} \right) \cdot \frac{dt}{dx}$.Using the quotient rule for $\frac{d}{dt} \left( \frac{g^{\prime}(t)}{f^{\prime}(t)} \right)$:$= \frac{f^{\prime}(t) \cdot g^{\prime \prime}(t) - g^{\prime}(t) \cdot f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^2} \cdot \frac{1}{f^{\prime}(t)}$.$= \frac{f^{\prime}(t) \cdot g^{\prime \prime}(t) - g^{\prime}(t) \cdot f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^3}$.

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

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