Let y = sqrt{p x^3 y}. If frac{dy}{dx} = frac | vedclass

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(A) Given $y = \sqrt{p x^3 y}$. Squaring both sides, we get $y^2 = p x^3 y$. Since $y 
eq 0$, we can divide by $y$ to get $y = p x^3$.Differentiating

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$1$

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(A) Given $y = \sqrt{p x^3 y}$. Squaring both sides, we get $y^2 = p x^3 y$. Since $y 
eq 0$, we can divide by $y$ to get $y = p x^3$.Differentiating both sides with respect to $x$, we get $\frac{dy}{dx} = 3 p x^2$.Given $\frac{dy}{dx} = \frac{3}{2}$ when $y = 1$. From $y = p x^3$, when $y = 1$, we have $1 = p x^3$, so $x^3 = \frac{1}{p}$, which means $x = p^{-1/3}$.Substitute $\frac{dy}{dx} = \frac{3}{2}$ and $x = p^{-1/3}$ into the derivative equation:$\frac{3}{2} = 3 p (p^{-1/3})^2$$\frac{3}{2} = 3 p (p^{-2/3})$$\frac{3}{2} = 3 p^{1/3}$$\frac{1}{2} = p^{1/3}$$p = (\frac{1}{2})^3 = \frac{1}{8}$.Note: The provided options do not contain the correct value $\frac{1}{8}$. Assuming a potential typo in the question where $y^2 = p x^3$ was intended, if $y^2 = p x^3$, then $2y \frac{dy}{dx} = 3 p x^2$. At $y=1$, $2(1)(\frac{3}{2}) = 3 p x^2 \implies 3 = 3 p x^2 \implies p x^2 = 1$. Also $1 = p x^3 \implies x = 1/p$. Substituting $x=1/p$ into $p x^2 = 1$ gives $p(1/p^2) = 1 \implies 1/p = 1 \implies p = 1$.

Let $y = \sqrt{p x^3 y}$. If $\frac{dy}{dx} = \frac{3}{2}$ when $y = 1$, then the value of $p$ is equal to

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