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If $\mathrm{x}=\log \mathrm{p}$ and $\mathrm{y}=\frac{1}{p}$ then $\frac{d y}{d x}=$
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$-\mathrm{e}^{-\mathrm{x}}$
$\begin{aligned} & x=\log p, y=\frac{1}{p} \\ \Rightarrow \quad & \frac{d y}{d p}=\frac{-1}{p^2} \Rightarrow \frac{d x}{d p}=\frac{1}{p} \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{d y / d p}{d x / d p}=\frac{-1}{p^2} \times p=\frac{-1}{p} \\ \because \quad & x=\log p \Rightarrow p=e^x\end{aligned}$
$\therefore \quad \frac{d y}{d x}=\frac{-1}{e^x}=-e^{-x}$.
$\therefore \quad \frac{d y}{d x}=\frac{-1}{e^x}=-e^{-x}$.
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