Finding the Derivative of a Logarithmic Function with a Variable Exponent

[lamerali]

Another problem I'm not sure of :(

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

my answer

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

x2y [tex]\frac{dy}{dx}[/tex] + [tex]\frac{dy}{dx}[/tex] - 4[tex]\frac{dy}{dx}[/tex] = -y [tex]^{2}[/tex] - lnx

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

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{-y ^{2} - lnx}{x2y - 3}[/tex]

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 [tex]\frac{dy}{dx}[/tex] for the function xy[tex]^{2}[/tex] + x lnx = 4y

my answer

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

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 = [tex]\frac{-y^{2} - lnx - 1}{ 2yx - 4}[/tex]

 

[Defennder]

Yeah that looks correct.

 

[lamerali]

Thank you :D