Implicit Differentiation

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nasi112

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Is it correct to say [MATH]y = f(x, y)[/MATH], then use implicit differentiation?
 
The question is strange. It would make a lot more sense if the f(x,y) was a constant.

The way it's written above x and y are independent variables and there should be no relation between them at all.
 
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)
 
nasi112 said:
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)
Click to expand...
How did you arrive at that expression?

Please share your work.
 
Subhotosh Khan said:
Post the EXACT and COMPLETE problem
Click to expand...

This is the exact problem.

Here how I arrived to [MATH]\frac{dy}{dx}[/MATH]
the total differential for [MATH]f[/MATH] is

[MATH]df = f_x \ dx + f_y \ dy[/MATH]
let [MATH]f_x = g(x, y)[/MATH] and [MATH]f_y = h(x, y)[/MATH]
then

[MATH]df = g(x, y) \ dx + h(x, y) \ dy[/MATH]
divide all sides by [MATH]dx[/MATH]
[MATH]\frac{df}{dx} = g(x, y) + h(x, y) \ \frac{dy}{dx}[/MATH]
then

[MATH]\frac{dy}{dx} = \frac{\frac{df}{dx} - g(x, y)}{h(x, y)}[/MATH]
 
nasi112 said:
I mean, how to proceed to the next step?
Click to expand...

As far as I can tell, the problem makes no sense at all as stated, and there is no reason to proceed at all.

Either the problem is just wrong, or there is some information in context (such as in instructions for a set of problems that say to take f(x,y) as a constant in each case) that would make it meaningful. I would ask your instructor whether the problem is misstated, and if not, what it means.
 
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]
 
Yeah, me too. I think that there is something wrong in this question. All the notes we have studied, there is no such question is asking for [MATH]\frac{dy}{dx}[/MATH] with implicit differentiation when [MATH]f(x,y)[/MATH] is presented.

I have sent an email to the professor to explain how to answer this problem. Or at least to tell if there is an error in the question.
 
Romsek said:
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]
Click to expand...

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.
 
nasi112 said:
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.
Click to expand...

that's only true if [MATH]y[/MATH] is not a function of [MATH]x[/MATH].

in which case [MATH]\dfrac{dy}{dx}=0[/MATH]
The whole problem is a mess.
 
Thanks all for the Try. I will wait for the professor reply and see what is the secret behind this mess.
 
If you insist on renaming f(x,y) you should call it z, certainly not y or x.
 
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