Solve for dy/dx if 2y^2-xy-x^2=0

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pope4

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I'm having some trouble understanding how to do implicit differentiation. I plugged this formula into an online calculator and it said that the answer is -2x-y, and I can't even begin to imagine where this came from.

I figured I could isolate for 2y^2 and take derivatives of both sides, but it still isn't making any sense. Taking 2y^2 as an example, I know that the derivative of y in respect to x is y' and, if I was dealing with x, I'd use the power rule, so how come (according to the calculator) is the answer 0?

I have no teacher for this course yet I've been understanding calc pretty well so far, but this is the one topic has be completely baffled. Thank you so much for any help!
 
pope4 said:
I'm having some trouble understanding how to do implicit differentiation. I plugged this formula into an online calculator and it said that the answer is -2x-y, and I can't even begin to imagine where this came from.

I figured I could isolate for 2y^2 and take derivatives of both sides, but it still isn't making any sense. Taking 2y^2 as an example, I know that the derivative of y in respect to x is y' and, if I was dealing with x, I'd use the power rule, so how come (according to the calculator) is the answer 0?

I have no teacher for this course yet I've been understanding calc pretty well so far, but this is the one topic has be completely baffled. Thank you so much for any help!
Click to expand...
Do you not have a textbook? That should tell you what to do. If you don't., try this one (or search for others):

Calculus I - Implicit Differentiation

In this section we will discuss implicit differentiation. Not every function can be explicitly written in terms of the independent variable, e.g. y = f(x) and yet we will still need to know what f'(x) is. Implicit differentiation will allow us to find the derivative in these cases. Knowing...
tutorial.math.lamar.edu

Now, I don't get -2x-y, so I don't know where that came from either! Probably you entered the wrong thing into the program, or it doesn't do what you think it does (which one is it?).

Please read up on the method, then try solving your problem and write back showing your work. There are some interesting things to discuss here.
 
Dr.Peterson said:
Do you not have a textbook? That should tell you what to do. If you don't., try this one (or search for others):

Calculus I - Implicit Differentiation

In this section we will discuss implicit differentiation. Not every function can be explicitly written in terms of the independent variable, e.g. y = f(x) and yet we will still need to know what f'(x) is. Implicit differentiation will allow us to find the derivative in these cases. Knowing...
tutorial.math.lamar.edu

Now, I don't get -2x-y, so I don't know where that came from either! Probably you entered the wrong thing into the program, or it doesn't do what you think it does (which one is it?).

Please read up on the method, then try solving your problem and write back showing your work. There are some interesting things to discuss here.
Click to expand...
I used Wolfram alpha, and I'm sure I must've typed something wrong and it spat out something I was not looking for!

I solved it and ended up getting y'=-((y+2x)/(x-4y)...?

I solved for each derivative separately and these are the results I got:
d/dx(2y^2)=d/dx(2yy)=(2dy/dx(y))+dy/dx(2y)=2y'y+2yy'=4yy'
d/dx(-xy)=-y+y'(-x)
d/dx(-x^2)=-2x
Then of course d/dx(0)=0

Then I plugged in the values into the equation:
4yy'+y-y'(-x)+2x=0
Then I went ahead and isolated for y', which left me with the equation I mentioned above.
 
pope4 said:
I used Wolfram alpha, and I'm sure I must've typed something wrong and it spat out something I was not looking for!

I solved it and ended up getting y'=-((y+2x)/(x-4y)...?

I solved for each derivative separately and these are the results I got:
d/dx(2y^2)=d/dx(2yy)=(2dy/dx(y))+dy/dx(2y)=2y'y+2yy'=4yy'
d/dx(-xy)=-y+y'(-x)
d/dx(-x^2)=-2x
Then of course d/dx(0)=0

Then I plugged in the values into the equation:
4yy'+y-y'(-x)+2x=0
Then I went ahead and isolated for y', which left me with the equation I mentioned above.
Click to expand...
Yes, that's right. When you differentiate the equation 2y^2-xy-x^2=0 with respect to x, you get 4yy'+y+xy'+2x=0, and solving for y', you get y'=(y+2x)/(4y-x).

Now, for fun, you might try solving the original equation for x, and differentiating that. Compare your result to the result of implicit differentiation.

When you've done all that, graph the equation and think about it again!
 
Yes, well the calculator gave you the wrong answer. It interpreted the question as asking for the derivative with respect to x given that y is a constant. I am not sure why the calculator decided that was what you were asking for, but the resulting answer has NOTHING to do with implicit differentiation where y is a function of x. Implicit differentiation is just a quick way to use the chain rule.

[math]\text {Set } z(x) = 2y^2 - xy - x^2.\\ \text {But it is given that } 2y^2 - xy - x^2 = 0.\\ \therefore \ z(x) = 0 \implies z’(x) = 0.[/math]
Obvious, right?

Now notice that z(x) is the sum of three functions of x. So, by the addition rule,

[math]z’(x) = \dfrac{d}{dx} z = \dfrac{d}{dx} 2y^2 + \dfrac{d}{dx} (-xy) + \dfrac{d}{dx} (-x^2).\\ \dfrac{d}{dx} 2y^2 = \left (\dfrac{d}{dy} 2y^2 \right ) * \dfrac{dy}{dx} = 4y * y’ \text { by power and chain rules.}\\ \dfrac{d}{dx} (-xy) = - x * \dfrac{dy}{dx} - 1 * y = - x * y’ - y \text { by multiplication and chain rule.}\\ \dfrac{d}{dx} (-x^2) = - 2x \text { by power rule.}\\ \therefore z’(x) = 4y * y’ - x * y’ - y - 2x = y’(4y - x) - y - 2x.\\ \text {But } z’(x) = 0 \implies y’(4y - x) - y - 2x = 0 \implies y’(4y - x) = y + 2x \implies \\ y’ = \dfrac{y + 2x}{4y - x}.[/math]
Implicit differentiation skips a lot of steps, but it essentially is just differentiating functions of y (where y is a function of x) using the chain rule.
 
Dr.Peterson said:
Now, for fun, you might try solving the original equation for x, and differentiating that. Compare your result to the result of implicit differentiation.
Click to expand...
I solved for y because we want y to check our result for y’.

[math]2y^2 - xy - x^2 = 0 \implies\\ y = \dfrac{x \pm \sqrt{(-x)^2 - 4(2)(-x^2)}}{2 * 2} = \dfrac{x \pm 3x}{4} \implies \\ y = x \text { or } y = - 0.5x \implies \\ y’ = 1 \text { or } y’ = - 0.5.\\ \text {CHECK}\\ y = x \implies 2y^2 - xy - x^2 = 2x^2 - x^2 - x^2 = 0. \ \checkmark \\ y’ = 1 \text { and } \dfrac{y + 2x}{4y - x} = \dfrac{x + 2x}{4x - x} = 1. \ \checkmark \\ y = -0.5x \implies 2y^2 - xy - x^2 = 2(0.25x^2) - (x)(-0.5x) - x^2 = 0.5x^2 + 0.5x^2 - x^2 = 0. \ \checkmark \\ y’ = - 0.5 \text { and } \dfrac{y + 2x}{4y - x} = \dfrac{-0.5x + 2x}{4(-0.5x) - x} = \dfrac{1.5x}{-2x - x} = -0.5. \ \checkmark [/math]
 
NotJeffM said:
I solved for y because we want y to check our result for y’.

[math]2y^2 - xy - x^2 = 0 \implies\\ y = \dfrac{x \pm \sqrt{(-x)^2 - 4(2)(-x^2)}}{2 * 2} = \dfrac{x \pm 3x}{4} \implies \\ y = x \text { or } y = - 0.5x \implies \\ y’ = 1 \text { or } y’ = - 0.5.\\ \text {CHECK}\\ y = x \implies 2y^2 - xy - x^2 = 2x^2 - x^2 - x^2 = 0. \ \checkmark \\ y’ = 1 \text { and } \dfrac{y + 2x}{4y - x} = \dfrac{x + 2x}{4x - x} = 1. \ \checkmark \\ y = -0.5x \implies 2y^2 - xy - x^2 = 2(0.25x^2) - (x)(-0.5x) - x^2 = 0.5x^2 + 0.5x^2 - x^2 = 0. \ \checkmark \\ y’ = - 0.5 \text { and } \dfrac{y + 2x}{4y - x} = \dfrac{-0.5x + 2x}{4(-0.5x) - x} = \dfrac{1.5x}{-2x - x} = -0.5. \ \checkmark [/math]
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Which, of course, is what I suggested the OP do. in order to discover this for himself.

It turns out that the equation can be solved by factoring (though I didn't try that at first because I didn't expect it to work); we get (2y+x)(y-x)=0, so y=-x/2 or y=x.

And if you graph the original equation on Desmos, you get this:

1683376507382.png​

A point that this can bring out is that while the result of implicit differentiation (a rational function) looks totally different from the actual derivative (a pair of constants), it applies only to points on the graph, where it does simplify to those two values.

In more typical examples, the graph is not so simple, but the same fact is true. And when you can't solve for y, it's much harder to show that your answer agrees with any other answer. Implicit differentiation can result in very different answers when you manipulate the equation before differentiating.

blamocur said:
Since nitpicking is my middle name: you get the above after you swap some of the signs :) To be fair, you did get the final result right.
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Yes, I copied the OP's work at that step and made small changes to it, forgetting to compare it to my own work, which had the right signs. I made the mistake of thinking that if someone's answer is right, the work must have been right. So this "nit" jumped onto me; it wasn't native.
 
Dr.Peterson said:
Which, of course, is what I suggested the OP do. in order to discover this for himself.

It turns out that the equation can be solved by factoring (though I didn't try that at first because I didn't expect it to work); we get (2y+x)(y-x)=0, so y=-x/2 or y=x.
Click to expand...
Yes, I noticed it could be factored as well, it was one of the first things I noticed since it was the only thing I recognized, but then I couldn't figure out what to do from there and ended up ignoring it until I had to use the factored form in the second part of the question. How would one go about solving the question with the 2 values for y though?
Dr.Peterson said:
Yes, I copied the OP's work at that step and made small changes to it, forgetting to compare it to my own work, which had the right signs. I made the mistake of thinking that if someone's answer is right, the work must have been right. So this "nit" jumped onto me; it wasn't native.
Click to expand...
Later on I went over my work again and noticed I switched some of the signs by accident and the mistake eluded me as I was tiptoeing around my own confusion. I went back and fixed it!
 
pope4 said:
Yes, I noticed it could be factored as well, it was one of the first things I noticed since it was the only thing I recognized, but then I couldn't figure out what to do from there and ended up ignoring it until I had to use the factored form in the second part of the question.
Click to expand...
There's a second part of the question:? Why didn't you show us the whole thing??

Factoring is not appropriate for implicit differentiation; I suppose the second part must tell you to solve and differentiate directly, as I suggested to do as an extra learning project.
pope4 said:
How would one go about solving the question with the 2 values for y though?
Click to expand...
That doesn't matter at all for implicit differentiation. But we've shown you all the work already!

Please do as we ask, and show the entire problem, so we know what you have to do:

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