Find y' at (0,1): Partial Derivative at (x,y)=(0,1)

[beaf123]

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]

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.

∂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.

 

[beaf123]

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]

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'.

 

[beaf123]

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′ = F^{`</sup><sub>1</sub> (x, y) / F<sup>`}₂ (x, y)
F`₂ (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]

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′ = F^{`</sup><sub>1</sub> (x, y) / F<sup>`}₂ (x, y)
F`₂ (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:

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

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.

∂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.

 

[beaf123]

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]

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.