If $x = t^{1/2}$, $y = t^{3/2}$, then $\frac{dy}{dx}$ =

Given $x = t^{1/2}$ and $y = t^{3/2}$, find $\frac{dy}{dx}$.

When both $x$ and $y$ depend on $t$:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

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For $x = t^{1/2}$:

$\frac{dx}{dt} = \frac{1}{2} \cdot t^{1/2 - 1}$

$\frac{dx}{dt} = \frac{1}{2}t^{-1/2}$

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For $y = t^{3/2}$:

$\frac{dy}{dt} = \frac{3}{2} \cdot t^{3/2 - 1}$

$\frac{dy}{dt} = \frac{3}{2}t^{1/2}$

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$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

$\frac{dy}{dx} = \frac{\frac{3}{2}t^{1/2}}{\frac{1}{2}t^{-1/2}}$

$\frac{dy}{dx} = \frac{3}{2}t^{1/2} \times \frac{2}{t^{-1/2}}$

$\frac{dy}{dx} = \frac{3}{2} \times 2 \times t^{1/2} \times t^{1/2}$

$\frac{dy}{dx} = 3 \times t^{1/2 + 1/2}$

$\frac{dy}{dx} = 3 \times t^{1}$

$\frac{dy}{dx} = 3t$

Therefore, $\frac{dy}{dx} = 3t$.