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find y' if x^(2y)=y^(3x)
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Take the total derivative of x^(2 y) =y^(3 x)\[x^{2 y} \left(\frac{2 y \text{Dt}[x]}{x}+2 \text{Dt}[y] \text{Log}[x]\right)=y^{3 x} \left(\frac{3 x \text{Dt}[y]}{y}+3 \text{Dt}[x] \text{Log}[y]\right)\]Solve the above for Dt[y].\[\text{Dt}[y]=\frac{y \text{Dt}[x] \left(2 x^{2 y} y-3 x y^{3 x} \text{Log}[y]\right)}{x \left(3 x y^{3 x}-2 x^{2 y} y \text{Log}[x]\right)} \]Divide each side by Dt[x]\[\frac{\text{Dt}[y]}{ \text{Dt}[x]}=\frac{y \left(2 x^{2 y} y-3 x y^{3 x} \text{Log}[y]\right)}{x \left(3 x y^{3 x}-2 x^{2 y} y \text{Log}[x]\right)} \]
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