If sqrt[3]{y} sqrt{x} = sqrt[6]{(x+y)^{5}},th | vedclass

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(B) Given the equation: $\sqrt[3]{y} \sqrt{x} = \sqrt[6]{(x+y)^{5}}$Raising both sides to the power of $6$,we get:$(y^{1/3} x^{1/2})^6 = ((x+y)^{5/6})

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$\frac{y}{x}$

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(B) Given the equation: $\sqrt[3]{y} \sqrt{x} = \sqrt[6]{(x+y)^{5}}$Raising both sides to the power of $6$,we get:$(y^{1/3} x^{1/2})^6 = ((x+y)^{5/6})^6$$y^2 x^3 = (x+y)^5$Let $y = vx$,then $dy = v dx + x dv$. Substituting this into the equation:$(vx)^2 x^3 = (x + vx)^5$$v^2 x^5 = x^5(1+v)^5$$v^2 = (1+v)^5$This implies $v$ is a constant,which is not possible for a general function. However,for homogeneous functions of degree $n$,if $f(y/x) = c$,then $\frac{dy}{dx} = \frac{y}{x}$.Alternatively,differentiating $y^2 x^3 = (x+y)^5$ implicitly:$2y y' x^3 + 3x^2 y^2 = 5(x+y)^4 (1+y')$Substitute $(x+y)^5 = y^2 x^3$,so $(x+y)^4 = \frac{y^2 x^3}{x+y}$:$2y y' x^3 + 3x^2 y^2 = 5 \frac{y^2 x^3}{x+y} (1+y')$Dividing by $x^2 y$:$2x y' + 3y = \frac{5xy}{x+y} (1+y')$$(2x y' + 3y)(x+y) = 5xy + 5xy y'$$2x^2 y' + 2xy y' + 3xy + 3y^2 = 5xy + 5xy y'$$y'(2x^2 + 2xy - 5xy) = 5xy - 3xy - 3y^2$$y'(2x^2 - 3xy) = 2xy - 3y^2$$y' x(2x - 3y) = y(2x - 3y)$Since $2x - 3y 
eq 0$,we get $\frac{dy}{dx} = \frac{y}{x}$.

If $\sqrt[3]{y} \sqrt{x} = \sqrt[6]{(x+y)^{5}}$,then $\frac{dy}{dx} = $

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