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

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(C) Given $x = \log p$,we can write $p = e^x$.Since $y = \frac{1}{p}$,substituting $p = e^x$ gives $y = \frac{1}{e^x} = e^{-x}$.Now,differentiate $y$

Vedclass · Accepted Answer

$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 0$

Vedclass · Answer

(C) Given $x = \log p$,we can write $p = e^x$.Since $y = \frac{1}{p}$,substituting $p = e^x$ gives $y = \frac{1}{e^x} = e^{-x}$.Now,differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(e^{-x}) = -e^{-x}$.Next,find the second derivative:$\frac{d^2y}{dx^2} = \frac{d}{dx}(-e^{-x}) = -(-e^{-x}) = e^{-x}$.Adding the first and second derivatives:$\frac{d^2y}{dx^2} + \frac{dy}{dx} = e^{-x} + (-e^{-x}) = 0$.Thus,the correct option is $C$.

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

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