# Partial derivative

x2y2 + (y+1)e-x=2 + x

Defines y as a differentiable function of x at point (x, y) = (0,1)

Find y′:

My attempt:

∂y/∂x =2xy3 + (-y-1)e-x=1

∂y/∂y = 3x2y2 - e-x=0

Plugging in for x and y ⇒

∂y/∂x = -3

∂y/∂x = -1

For some reason I think y′ is defined as
(∂y/∂x) /(∂y/∂y) = 3

At leas this give the right answer (from solutions), but I am uncertain so I hope someone can correct me where I am wrong.

Mark44
Mentor
In future posts, please do not delete the homework template.

x2y2 + (y+1)e-x=2 + x

Defines y as a differentiable function of x at point (x, y) = (0,1)

Find y′:

My attempt:

∂y/∂x =2xy3 + (-y-1)e-x=1
No, not even close.
First, you shouldn't be using partial derivatives, as the problem statement says that y is a differentiable function of x.
Second, and more important, to find dy/dx, you need to use implicit differentiation.
beaf123 said:
∂y/∂y = 3x2y2 - e-x=0

Plugging in for x and y ⇒

∂y/∂x = -3

∂y/∂x = -1

For some reason I think y′ is defined as
(∂y/∂x) /(∂y/∂y) = 3

At leas this give the right answer (from solutions), but I am uncertain so I hope someone can correct me where I am wrong.

I looked up implicit differentiation in my book ( thanks for pointing me in the right direction), but it seems like that is what I have been doing even without knowing it. What I did was correct and the answer is correct.

Mark44
Mentor
I looked up implicit differentiation in my book ( thanks for pointing me in the right direction), but it seems like that is what I have been doing even without knowing it. What I did was correct and the answer is correct.
The answer might possibly be correct, but your work is not correct.

Starting with this equation:
##x^2y^2 + (y + 1)e^{-x} = 2 + x##
and differentiating implicitly with respect to x, we get
##2xy^2 + 2x^2y*y' - (y + 1)e^{-x} + y' e^{-x} = 1##
To continue, put all of the terms involving y' on one side, and all the others on the other side, and solve algebraically for y'.

The formula I found in the book says:

Slope of a level curve ( which is what I have since 2 is a constant)

F(x,y) = c ⇒ y′ = F1 (x, y) / F2 (x, y)
F2 (x, y) ≠ 0

which is exactly what I did. Maybe you would get the same answer doing it your way, but what I did is correct as well.

Mark44
Mentor
The formula I found in the book says:

Slope of a level curve ( which is what I have since 2 is a constant)

F(x,y) = c ⇒ y′ = F1 (x, y) / F2 (x, y)
F2 (x, y) ≠ 0

which is exactly what I did. Maybe you would get the same answer doing it your way, but what I did is correct as well.
Here's what you posted earlier:
beaf123 said:
x2y2 + (y+1)e-x=2 + x

Defines y as a differentiable function of x at point (x, y) = (0,1)
The function for your level curve would be F(x, y) = C, or x2y2 + (y+1)e-x - x =2
So the left side of the last equation if F(x, y), and C = 2
beaf123 said:
Find y′:

My attempt:

∂y/∂x =2xy3 + (-y-1)e-x=1
2xy3 can't be right.
Also, the above should be ∂F/∂x and below should be ∂F/∂y, since ∂y/∂y is 1.
beaf123 said:
∂y/∂y = 3x2y2 - e-x=0

I think you stumbled onto the correct answer by accident, which is not a good way to do things. You lucked out that the point in question is (0, 1), so lots of terms drop out.

You are right that it should be ∂F/∂x and below should be ∂F/∂y, since ∂y/∂y is 1 and also I wrote y3 instead of y2 which could be confusing. But as you said it doesent matter cus it drops out.

Ray Vickson
Homework Helper
Dearly Missed
You are right that it should be ∂F/∂x and below should be ∂F/∂y, since ∂y/∂y is 1 and also I wrote y3 instead of y2 which could be confusing. But as you said it doesent matter cus it drops out.

I hope you realize that it matters a lot. You would really see the difference if you started at a different point, such as ##(x,y) = (1,y_1)##, where
$$y_1 = \frac{e^{-1}}{2} \left( -1 + \sqrt{12e^2 - 4e +1} \right).$$
Another point is that you might be marked wrong even if your answer is correct but your argument and formulas are incorrect.

• Mark44