If x = log p and y = frac{1}{p}, then frac{dy | vedclass

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(A) Given that $x = \log p$ and $y = \frac{1}{p}$.First, differentiate $y$ with respect to $p$:$\frac{dy}{dp} = \frac{d}{dp}(p^{-1}) = -p^{-2} = -\fra

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$-e^{-x}$

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(A) Given that $x = \log p$ and $y = \frac{1}{p}$.First, differentiate $y$ with respect to $p$:$\frac{dy}{dp} = \frac{d}{dp}(p^{-1}) = -p^{-2} = -\frac{1}{p^2}$.Next, differentiate $x$ with respect to $p$:$\frac{dx}{dp} = \frac{d}{dp}(\log p) = \frac{1}{p}$.Using the chain rule for parametric differentiation:$\frac{dy}{dx} = \frac{dy/dp}{dx/dp} = \frac{-1/p^2}{1/p} = -\frac{1}{p}$.Since $x = \log p$, we have $p = e^x$.Substituting $p = e^x$ into the expression for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{1}{e^x} = -e^{-x}$.

If $x = \log p$ and $y = \frac{1}{p}$, then $\frac{dy}{dx} = $

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