Implicit Differentiation

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  • #1
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Homework Statement


[itex]\dfrac{x^2}{x+y}=y^2+8[/itex]

Homework Equations



Quotient Rule: [itex]\dfrac{g(x)\cdot f'(x)-g'(x)\cdot f(x)}{(g(x))^2}[/itex]

Product Rule: [itex]f(x)\cdot g'(x)+g(x)\cdot f'(x)[/itex]

The Attempt at a Solution



[itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

I feel like there are a couple of ways to go about this. Would it be easier to flip the denominator and use the product rule? I just used the quotient from here on out.

What you see my trying to do is get [itex]\dfrac{dy}{dx}[/itex] on the left side and everything else on the right, then factor [itex]\dfrac{dy}{dx}[/itex] out. This is my first encounter with this type of problem so I get confused very fast.

[itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

[itex]\dfrac{(2x^2+2xy\cdot\dfrac{dy}{dx})-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

[itex]\dfrac{\dfrac{dy}{dx}-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{(2x^2+2xy\cdot\dfrac{dy}{dx})}[/itex]

[itex]\dfrac{-1}{x^2}\cdot\dfrac{\dfrac{dy}{dx}-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{(2x^2+2xy)}\cdot\dfrac{-1}{x^2}[/itex]

[itex]\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{-x^2(2x^2+2xy)}[/itex]

[itex]\dfrac{1}{\dfrac{dy}{dx}}\cdot\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{-x^2(2x^2+2xy)}\cdot\dfrac{1}{\dfrac{dy}{dx}}[/itex]

[itex]\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{\dfrac{dy}{dx}\cdot(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y}{-x^2(2x^2+2xy)}[/itex]

I don't want to go much farther because I could be doing this wrong. On the left side, I want to factor, but I'm curious if this is right so far or have I made any errors?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


[itex]\dfrac{x^2}{x+y}=y^2+8[/itex]

Homework Equations



Quotient Rule: [itex]\dfrac{g(x)\cdot f'(x)-g'(x)\cdot f(x)}{(g(x))^2}[/itex]

Product Rule: [itex]f(x)\cdot g'(x)+g(x)\cdot f'(x)[/itex]

The Attempt at a Solution



[itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

I don't want to go much farther because I could be doing this wrong. On the left side, I want to factor, but I'm curious if this is right so far or have I made any errors?

Where does the ##y y'## come from on the left?
 
  • #3
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Where does the yy′yy′y y' come from on the left?

Are you talking about where I have [itex](x+y\cdot \dfrac{dy}{dx})[/itex] or [itex]-(1\cdot \dfrac{dy}{dx})[/itex]?
 
  • #4
ehild
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Homework Statement


[itex]\dfrac{x^2}{x+y}=y^2+8[/itex]

Homework Equations



Quotient Rule: [itex]\dfrac{g(x)\cdot f'(x)-g'(x)\cdot f(x)}{(g(x))^2}[/itex]

Product Rule: [itex]f(x)\cdot g'(x)+g(x)\cdot f'(x)[/itex]

The Attempt at a Solution



[itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

The differentiation of the left side is wrong. What are f and g ? Just apply the Quotient Rule properly.
 
  • #5
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The differentiation of the left side is wrong. What are f and g ? Just apply the Quotient rule properly.

Oops, here they are:
[itex]f(x)=x^2[/itex]
[itex]f'(x)=2x[/itex]
[itex]g(x)=(x+y)[/itex]
[itex]g'(x)=(x+y\cdot\dfrac{dy}{dx})[/itex]

Edit: Oh, I need it to say [itex]\dfrac{(x+y)(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]
 
  • #6
ehild
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Oops, here they are:
[itex]f(x)=x^2[/itex]
[itex]f'(x)=2x[/itex]
[itex]g(x)=(x+y)[/itex]
[itex]g'(x)=(x+y\cdot\dfrac{dy}{dx})[/itex]
The last equation is wrong. The derivative of a sum is the sum of derivatives. What is dx/dx? and dy/dx is not yy'.
 
  • #7
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Would it just be [itex]\dfrac{dy}{dx}[/itex]
 
  • #8
ehild
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Oops, here they are:
[itex]f(x)=x^2[/itex]
[itex]f'(x)=2x[/itex]
[itex]g(x)=(x+y)[/itex]
[itex]g'(x)=(x+y\cdot\dfrac{dy}{dx})[/itex]

Edit: Oh, I need it to say [itex]\dfrac{(x+y)(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]
Still wrong. What should be the denominator? Is not it g2?
 
  • #9
ehild
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Would it just be [itex]\dfrac{dy}{dx}[/itex]
Of course.
 
  • #11
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I started moving the [itex]\dfrac{dy}{dx}[/itex] to the right side instead, but didn't finish just yet.

[itex]\dfrac{(x+y)(2x)-(\dfrac{dy}{dx})(x^2)}{(x+y^2)^2}=2y\cdot\dfrac{dy}{dx}[/itex]

[itex]\dfrac{1}{\dfrac{dy}{dx}}\cdot\dfrac{2x^2+2xy-x^2\cdot\dfrac{dy}{dx}}{(x+y^2\cdot\dfrac{dy}{dx})^2}=2y\cdot\dfrac{dy}{dx}\cdot\dfrac{1}{\dfrac{dy}{dx}}[/itex]

[itex]\dfrac{1}{2y}\cdot\dfrac{2x^2+2xy-x^2}{(x+y^2\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{\dfrac{dy}{dx}}\cdot\dfrac{1}{2y}[/itex]

[itex]\dfrac{2x^2+2xy-x^2}{2y(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{\dfrac{dy}{dx}}{\dfrac{dy}{dx}}[/itex]
 
  • #12
Ray Vickson
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Are you talking about where I have [itex](x+y\cdot \dfrac{dy}{dx})[/itex] or [itex]-(1\cdot \dfrac{dy}{dx})[/itex]?

I am talking about the ##y \frac{dy}{dx}## part.
 
  • #13
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I am talking about the ydydxydydxy \frac{dy}{dx} part.

That was my fault. It was supposed to be [itex](x+y)[/itex]
 
  • #14
ehild
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I started moving the [itex]\dfrac{dy}{dx}[/itex] to the right side instead, but didn't finish just yet.

[itex]\dfrac{(x+y)(2x)-(\dfrac{dy}{dx})(x^2)}{(x+y^2)^2}=2y\cdot\dfrac{dy}{dx}[/itex]
Wrong denominator on the left side.
 
  • #15
ehild
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You confuse yourself when dy/dx appears on both sides. It has no sense dividing by dy/dx.
Multiply the original equation by (x+y), then you need to differentiate the equation ##x^2=(x+y)(y^2+8)##.
 
  • #16
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You confuse yourself when dy/dx appears on both sides. It has no sense dividing by dy/dx.

I fixed that after that, it should've been like the rest.

Multiply the original equation by (x+y), then you need to differentiate the equation x2=(x+y)(y2+8)x2=(x+y)(y2+8)x^2=(x+y)(y^2+8).

Are you saying that after getting rid of that (x+y) in the denominator, I would distribute on the right side and use the product rule?

See this is my problem in math, I follow the pattern I first learn from when introduced into something new. When you said to multiply the original equation by (x+y), I wouldn't have thought of that because I've been following what I've been previously doing, by using the quotient rule because there's a fraction on the left side.

I want to clarify on something, after doing this:

[itex]x^2=(x+y)(y^2+8)[/itex]

In the second step after distributing,

[itex]x^2=xy^2\cdot\dfrac{dy}{dx}+8x+y^3\cdot\dfrac{dy}{dx}+8y\cdot\dfrac{dy}{dx}[/itex]

When I get here, am I using the chain rule on xy^2 and the product rule on 8x and 8y?
 
  • #17
ehild
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I fixed that after that, it should've been like the rest.
The rest was also wrong. There is no derivative in the denominator on the left side. Why did you put dy/dx there again? And I do not follow what you did.


Are you saying that after getting rid of that (x+y) in the denominator, I would distribute on the right side and use the product rule?
No need to distribute. Use the product rule, but differentiate both sides!
See this is my problem in math, I follow the pattern I first learn from when introduced into something new. When you said to multiply the original equation by (x+y), I wouldn't have thought of that because I've been following what I've been previously doing, by using the quotient rule because there's a fraction on the left side.

It is all right to differentiate the original equation, but you get a more complicate equation for dy/dx, and you looked confused. You need to use basic rules of multiplication and addition, instead of "following patterns".
the left side is x2. What is the derivative?
The right side is (x+y)(y2+8). Apply product rule to get the derivative with respect x.
 
  • #18
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The rest was also wrong. There is no derivative in the denominator on the left side. Why did you put dy/dx there again? And I do not follow what you did.

My professor was saying whenever there's a "y", you need to multiply by [itex]\dfrac{dy}{dx}[/itex]. He never showed us an example of a problem given a fraction so I don't know about that.

[itex]x^2=(x+y)(y^2+8)[/itex]

[itex]f(x)=x+y[/itex]

[itex]f'(x)=\dfrac{dy}{dx}[/itex]

[itex]g(x)=y^2+8[/itex]

[itex]g'(x)=2y\dfrac{dy}{dx}[/itex]

[itex]x^2=(x+y)(2y\cdot\dfrac{dy}{dx})+(y^2+8)(\dfrac{dy}{dx})[/itex]

[itex]x^2=2xy\cdot\dfrac{dy}{dx}+2y^2\cdot\dfrac{dy}{dx}+y^2\cdot\dfrac{dy}{dx}+8\cdot\dfrac{dy}{dx}[/itex]

Would you distribute like I just did on the last part there? Doesn't seem correct.
 
  • #19
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My professor was saying whenever there's a "y", you need to multiply by [itex]\dfrac{dy}{dx}[/itex].
You might have heard your professor incorrectly. When you're doing implicit differentiation, where there's a y, upon differentiation you get ##\frac{dy}{dx}##. You aren't multiplying by ##\frac{dy}{dx}##.

For example, if there's a term of ##y^2##, differentiating it gives ##\frac d{dx}\left(y^2\right) = 2y\cdot \frac {dy}{dx}##. That last factor comes from the chain rule. ##\frac d{dx}\left(y^2\right) = \frac d {dy} \left(y^2\right) \cdot \frac {dy}{dx}##.
 
  • #20
ehild
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My professor was saying whenever there's a "y", you need to multiply by [itex]\dfrac{dy}{dx}[/itex]. He never showed us an example of a problem given a fraction so I don't know about that.

[itex]x^2=(x+y)(y^2+8)[/itex]

[itex]f(x)=x+y[/itex]

[itex]f'(x)=\dfrac{dy}{dx}[/itex]

[itex]g(x)=y^2+8[/itex]

[itex]g'(x)=2y\dfrac{dy}{dx}[/itex]

[itex]x^2=(x+y)(2y\cdot\dfrac{dy}{dx})+(y^2+8)(\dfrac{dy}{dx})[/itex]

You have to differentiate both sides! so the left side is d(x2)/dx = ?
About the saying of your professor, whenever you differentiate a function of y, differentiate it with respect to y first, then multiply by y'. But in case of a fraction, it is f'g-fg' divided by the square of the original denominator.
 
  • #21
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I have a question, does [itex]y'[/itex] mean the same as [itex]y\cdot\dfrac{dy}{dx}[/itex]?
 
  • #22
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I have a question, does [itex]y'[/itex] mean the same as [itex]y\cdot\dfrac{dy}{dx}[/itex]?
Absolutely not. y' is alternate notation for ##\frac{dy}{dx}##. With the y' notation, the context should make clear that it is the derivative with respect to x.
 
  • #23
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Ok, thanks. I'm working on it.
 

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