# Logarithmic differentiation

1. Sep 29, 2008

### lamerali

Another problem i'm not sure of :(

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

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

2. Sep 29, 2008

### cristo

Staff Emeritus
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)

3. Sep 29, 2008

### lamerali

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

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

4. Sep 29, 2008

### Defennder

Yeah that looks correct.

5. Sep 29, 2008

Thank you :D