Derivative of y=f(y)

Jumblon wrote: »

Well, it should actually be y=f(x,y).

In this case you have to perform partial differentiation on the function f. If we write y = y(x) we get y(x) = f(x,y(x)). Then, differentiating both sides does something like

[LATEX]\displaystyle \frac{d}{dx}y(x) =\frac{d}{dx} f(x,y(x)) [/LATEX]
[LATEX]\displaystyle \frac{dy}{dx} =\frac{\partial f}{\partial x} +
\frac{\partial f}{\partial y}\frac{dy}{dx}[/LATEX]

so that

[LATEX]\displaystyle \frac{dy}{dx} =\frac{1}{1-\frac{\partial f}{\partial y}}\frac{\partial f}{\partial x} [/LATEX]

So, for instance, if y=sin(xy)x^2=f(x,y) then

[LATEX]\displaystyle \frac{\partial f}{\partial x} = \sin(xy) 2x + y \cos(xy) x^2; \;\; \frac{\partial f}{\partial y} = x \cos(xy) x^2 = \cos(xy) x^3[/LATEX]

and so

[LATEX]\displaystyle \frac{dy}{dx} =\frac{1}{1-\cos(xy) x^3} (\sin(xy) 2x + y \cos(xy) x^2 )[/LATEX]