NO!View attachment 24047
Is it correct to say [MATH]y = f(x, y)[/MATH], then use implicit differentiation?
What should I do to get rid of f(x,y)?
Post the EXACT and COMPLETE problemWhat should I do to get rid of f(x,y)?
How did you arrive at that expression?I can find the differential [MATH]df[/MATH]
then [MATH]\frac{dy}{dx} = \frac{\frac{df}{dx} - g(x, y)}{h(x, y)}[/MATH]
but even If I do that, it is not implicit differentiation as well as the answer is awkward (looks strange)
Post the EXACT and COMPLETE problem
but I don't have f(x). I have f(x,y).
yeah, bad choice of name... y = g(x), i.e. y is some function of x
I mean, how to proceed to the next step?
let's just call it [MATH]y(x)[/MATH]
[MATH]f(x,y) = x^3 y^2 + \ln(x^2y^5+x^2y^3) \\ \dfrac{\partial f}{\partial x} = 3x^2 y^2 + x^3 2yy^\prime + \dfrac{2x y^5 + 5x^2 y^4 y^\prime + 2x y^3 + 3x^2 y^2 y^\prime}{x^2 y^5 + x^2 y^3}[/MATH]
and solve for [MATH]y^\prime[/MATH]
Thanks Romsek for the hard work. I have learnt that when I take a partial derivative of [MATH]f[/MATH] with respect to [MATH]x[/MATH], I have to treat [MATH]y[/MATH] as a constant.
But I think that you mean, I have to replace [MATH]f(x, y)[/MATH] with [MATH]y[/MATH], and use implicit differentiation as if I have only one variable.
But is it the COMPLETE problem statement?This is the exact problem.