Logarithmic differentiation

[lamerali]

Another problem i'm not sure of :(

find \frac{dy}{dx} for the function xy^{2} + x lnx = 4y

my answer

y^{2} + x2y \frac{dy}{dx} + lnx + x (1/x) \frac{dy}{dx} = 4\frac{dy}{dx}

x2y \frac{dy}{dx} + \frac{dy}{dx} - 4\frac{dy}{dx} = -y ^{2} - lnx

\frac{dy}{dx} ( x2y - 3) = -y^{2} - lnx

\frac{dy}{dx} = \frac{-y ^{2} - lnx}{x2y - 3}

I'm not sure if this is the correct answer again any guidance is greatly appreciated!
Thank you

[cristo]

Another problem i'm not sure of :(

find \frac{dy}{dx} for the function xy^{2} + x lnx = 4y

my answer

y^{2} + x2y \frac{dy}{dx} + lnx + x (1/x) \frac{dy}{dx} = 4\frac{dy}{dx}

Why does the fourth term here have a dy/dx in it? The derivative of xln(x) wrt x is ln(x)+x(1/x)

[lamerali]

so should the dy/dx be eliminated from the ln(x) + x(1/x) completely? leaving the resulting derivative equal to

dy/dx = \frac{-y^{2} - lnx - 1}{ 2yx - 4}

[Defennder]

Yeah that looks correct.

[lamerali]

Thank you :D