If x = log t, t 0 and y = frac{1}{t},then f | vedclass

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(B) Given $x = \log t$ and $y = \frac{1}{t}$.First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.Since $y = t^{-1}$

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$-\frac{dy}{dx}$

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(B) Given $x = \log t$ and $y = \frac{1}{t}$.First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.Since $y = t^{-1}$,$\frac{dy}{dt} = -t^{-2} = -\frac{1}{t^2}$.Since $x = \log t$,$\frac{dx}{dt} = \frac{1}{t}$.Therefore,$\frac{dy}{dx} = \frac{-1/t^2}{1/t} = -\frac{1}{t} = -y$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2 y}{dx^2}$:$\frac{d^2 y}{dx^2} = \frac{d}{dx}(-y) = -\frac{dy}{dx}$.Substituting $\frac{dy}{dx} = -y$ into the expression:$\frac{d^2 y}{dx^2} = -(-y) = y$.Wait,let's re-evaluate: $\frac{d}{dx}(-y) = -\frac{dy}{dt} \cdot \frac{dt}{dx} = -(-t^{-2}) \cdot t = t^{-1} = y$.Thus,$\frac{d^2 y}{dx^2} = y$.

If $x = \log t, t > 0$ and $y = \frac{1}{t}$,then $\frac{d^2 y}{d x^2} =$

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