Implicit Differentiation Clarification

[Asphyxiated]

Homework Statement

$$2x^{2} - 3y^{2} = 4$$

Homework Equations

We say that y is an implicit function of x if we are given an equation:

$$\sigma(x,y) = \tau(x,y)$$

Then to differentiate we do:

$$\frac {d(\sigma(x,y))} {dx} = \frac {d(\tau(x,y))} {dx}$$

The Attempt at a Solution

$$2x^{2} - 3y^{2} = 4$$

$$\frac {d(2x^{2} - 3y^{2})} {dx} = \frac {d(4)} {dx}$$

$$4x - 6y \frac {dy} {dx} = 0$$

$$-6y \frac {dy} {dx} = -4x$$

$$\frac {dy} {dx} = \frac {4x} {6y} = \frac {2x} {3y}$$

If this is correct, that's great, but I really do not understand why the $$\frac {dy} {dx}$$ gets put into the equation. Of course I understand its function but I really don't understand the rules that go along with placing it in the equation. If I add it to one side shouldn't I have to add it to the other?

If I am wrong, please tell me what I did wrong and how to fix it.

Thanks!

 

[Dick]

That is great. Sure, dy/dx is what you want to find, isn't it? I don't see why you think you 'added it to one side'. You just differentiated and solved for it. Didn't you? There's nothing odd or wrong about that.

 

[Asphyxiated]

Well, I guess I mean after this:

$$\frac {d(2x^{2} - 3y^{2})} {dx} = \frac {d(4)} {dx}$$

you get:

$$4x - 6y \frac {dy} {dx} = 0$$

Where does the dy/dx come from? It's not leftover from the equation on the left side so I must have added it to the equation, and as far as I know, when you add (or subtract or anything) to/from one side you must also do it to the other. So that's where I am second guessing myself I suppose because it feels wrong to me to add it to the left and not the whole equation... if you get what I mean...

 

[Dick]

You didn't 'add it'. You differentiated d(y^2)/dx and got 2*y*dy/dx using the chain rule. Or I at least sincerely hope you did. I'm now guessing you didn't. But that's what you should have done.

 

[Asphyxiated]

um yeah I didn't, I mean I 'kind of' did, I took the derivative of $$2x^{2}-3y^{2}$$ and then added dy/dx to the result because that's how it was explained in my book (note it didn't say to "add dy/dx" I just got that from the examples.) If possible could you further explain what you just said?

 

[Dick]

Chain rule, man. f(y(x))'=f'(y(x))*y'(x). Look it up. f here is the squaring function. f(u)=u^2. So f'(u)=2u. f'(y(x))=2*y(x). y'(x)=dy/dx. Starting to make sense?

 

[Asphyxiated]

Yeah, I guess I just never carried the thought process that far... but I am pretty sure I understand you now. In my equation x is the independent variable and y depends on x so y is actually y(x), am I thinking about this correctly now?

 

[Dick]

Yeah, I guess I just never carried the thought process that far... but I am pretty sure I understand you now. In my equation x is the independent variable and y depends on x so y is actually y(x), am I thinking about this correctly now?

Right. So ((y(x))^2)'=2*y(x)*y'(x) where "'" is d/dx.

 

[Asphyxiated]

Hey, whenever you get online can you help me out with this question Dick? It's the same type of question from the same section, I think I am just a little confused:

$$y = \frac {1} {(x+y)}$$

$$\frac {d(y)} {dx} = \frac {d(\frac {1} {(x+y)})} {dx}$$

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} * 1 + \frac {dy} {dx}$$

At this point I would consider myself stuck, the only move I see is this:

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} + \frac {dy} {dx}$$

$$\frac {dy} {dx} - \frac {dy} {dx} = - \frac {1} {(x+y)^{2}}$$

and then you get 0 = which is not what I am trying to solve for obviously.. so I made a mistake some where, also I am not sure if

$$1 + \frac {dy} {dx}$$

in

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} * 1 + \frac {dy} {dx}$$

should be considered a binomial such as:

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} * (1 + \frac {dy} {dx})$$

Hopefully that's not all too confusing..., thanks for the help in advance!

______________________________________________________________________

Another question just to check to see if I understand this:

$$y^{2} + 3y - 5 = x$$

$$\frac {d(y^{2}+3y-5)} {dx} = \frac {d(x)} {dx}$$

$$2y \frac {dy} {dx} + 3 \frac {dy} {dx} = 1$$

$$\frac {dy} {dx}(2y +3) = 1$$

$$\frac {dy} {dx}(2y+3)-1 = 0$$

$$(2y+3)-1 = \frac {dy} {dx}$$

I am not really sure if the last move is legal because 0 / anything is 0 so... is it more like this:

$$\frac {dy} {dx}(2y +3) = 1$$

$$\frac {dy} {dx} = \frac {1} {(2y+3)}$$

 

[Dick]

I think your main problem is just that your algebra is a little shakey. If it were more solid, you would know what's right and what's wrong. For the second one, this is correct,
$$\frac {dy} {dx} = \frac {1} {(2y+3)}$$
for the first one this is correct,

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} * (1 + \frac {dy} {dx})$$.

Now you have to expand it and move all the dy/dx stuff to one side and everything else to the other side. The other stuff is just plain wrong algebraically, and as part of your algebra practice, you should tell me why they are wrong.

For example, why is going from here,

$$\frac {dy} {dx}(2y+3)-1 = 0$$

to here,

$$(2y+3)-1 = \frac {dy} {dx}$$

just plain dead wrong.

 

[Asphyxiated]

Well to answer the last question you posed, the best answer I can give you as to why its wrong is that 0 divided by anything is going to 0, not my dy/dx, maybe there is more that I forgot as to why it is not legal to make that move.

Also after expanding the term in the first problem I gave you I get:

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} * (1 + \frac {dy}{dx})$$

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} - \frac {1} {(x+y)^{2}} (\frac {dy}{dx})$$

but after this step if I divide to move the dx/dy on the right to the left it will leave me with 1 on the left which is not correct either...

Also you are quite right about my algebra skills, algebra II is the last class I took in high school and then started teaching myself trig, but there was about a 4 year gap in between algebra II and starting trig... but thanks for all the help man!

 

[Dick]

Ok, simpler question in the same form. Why don't x*a-1=0 and a-1=x have the same solution? That is essentially what you did with the question I posed. The simple answer to why it is illegal to make that move is that there is no legal way to make that rearrangement. Whatever is not legal is PROHIBITED. Besides, they give you different solutions for x. You have got to get those algebra skills up. Your math future depends on it.

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} - \frac {1} {(x+y)^{2}} (\frac {dy}{dx})$$

goes to,

$$\frac {dy} {dx} + \frac {1} {(x+y)^{2}} (\frac {dy}{dx})= - \frac {1} {(x+y)^{2}}$$

as a first step. Now can you solve for dy/dx?

 

[Asphyxiated]

not exactly no... I briefly explored that option before posting my last posting, I have 'explored' it more since you confirm that was the direction I was suppose to go in but things get convoluted quickly here for me...

$$\frac {dy} {dx} + \frac {1} {(x+y)^{2}} (\frac {dy} {dx}) = - \frac {1} {(x+y)^{2}}$$

$$\frac {dy} {dx} + \frac {dy} {dx} = \frac {- \frac {1} {(x+y)^{2}}} {\frac {1} {(x+y)^{2}}}$$

then the large fraction on right becomes -1 and i think you could go:

$$\frac {dy} {dx} (1+1) = -1$$

and I could ultimately end up with dy/dx = -1/2 but I have no x or y variables left and again must be doing it wrong... sorry I am not catching on here man

 

[Dick]

Ok. Your algebra, in fact, sucks. Big time. The left side becomes dy/dx*(1+1/(x+y)^2)). Just factor out the dy/dx. That is one of the few legal things you can do. That equals -1/(x+y)^2. Now you have an equation that is schematically a*(dy/dx)=b. Solve for dy/dx. And don't use voodoo.

 

[Asphyxiated]

Alright man, I got to say, I don't know how I missed that, I'm usually pretty comfortable with factoring...but anyway I think this is right then...

$$\frac {dy} {dx} (1 + \frac {1} {(x+y)^{2}}) = - \frac {1} {(x+y)^{2}}$$

$$\frac {dy} {dx} = \frac {- \frac {1} {(x+y)^{2}}} {(1 + \frac {1} {(x+y)^{2}})}$$

yay or nay?

 

[Dick]

Yay. You can simplify that a little bit. If you dare...

 

[Asphyxiated]

I'll give it a shot, maybe find some redemption, although it could go terribly wrong as well, haha

$$\frac {dy} {dx} = \frac {- \frac {1} {(x+y)^{2}}} {(1 + \frac {1} {(x+y)^{2}})}$$

$$\frac {dy} {dx} = -\frac {1} {(x+y)^{2}} * \frac {1 + (x+y)^{2}} {1}$$

$$\frac {dy} {dx} = - \frac {1 + (x+y)^{2}} {(x+y)^{2}}$$

$$\frac {dy} {dx} = - \frac {1} {(x+y)^{2}} + 1$$

that's what I got, I don't think there is any voodoo in there to speak of but as of right now, that confidence is kind of shaken

 

[Dick]

Way wrong. Here's a trick that might help. Replace (x+y) with a number, like 2. Now figure out the value of the right side of each line. Where did it change from -1/5, the starting value to 3/4 the ending value? How many other values are in between? Where are the bad steps?

 

[Asphyxiated]

ok how about this:

$$\frac {dy} {dx} = \frac {-\frac {1} {(x+y)^{2}}} {(1+\frac {1} {(x+y)^{2}})}$$

$$\frac {dy} {dx} = \frac {-\frac {1} {(x+y)^{2}}} {\frac {(x+y)^{2}+1} {(x+y)^{2}}}$$

$$\frac {dy} {dx} = -\frac {1} {(x+y)^{2}} * \frac {(x+y)^{2}} {(x+y)^{2}+1}$$

$$\frac {dy} {dx} = -\frac {(x+y)^{2}} {(x+y)^{2}((x+y)^{2} +1)}$$

$$\frac {dy} {dx} = -\frac {1} {(x+y)^{2} +1}$$

I checked for consistency in the results all the way through, so I think that this is correct... NO VOODOO

 

[Dick]

Now that's correct. No voodoo this time. If you put (x+y)=2, then the first line is -1/5 and the last line is -1/5. That's a pretty good check against voodoo, try checking numbers for a while, until you feel more confident.

 

[Asphyxiated]

ok, going for broke here Dick, here is another problem, I am rather certain that it is correct and simplified.

$$xy^{3}=y+x$$

$$\frac {d(xy^{3})} {dx} = \frac {d(y+x)} {dx}$$

$$x*3y^{2}\frac{dy}{dx}+y^{3}=\frac {dy}{dx}+1$$

$$x * 3y^{2}\frac{dy}{dx} = \frac {dy}{dx} + 1 - y^{3}$$

$$x * 3y^{2}\frac{dy}{dx} - \frac{dy}{dx} = 1 - y^{3}$$

$$3y^{2}\frac {dy}{dx}-\frac{dy}{dx} = \frac {1-y^{3}} {x}$$

$$\frac{dy}{dx}(3y^{2} - 1) = \frac {1-y^{3}} {x}$$

$$\frac {dy}{dx} = \frac {\frac {1-y^{3}}{x}} {3y^{2}-1}$$

then simplified:

$$\frac {dy}{dx} = \frac {1-y^{3}} {x} * \frac {1} {3y^{2}-1}$$

$$\frac {dy}{dx} = \frac {1-y^{3}} {(x)(3y^{2}-1)}$$

 

[Dick]

Goof alert here.

$$x * 3y^{2}\frac{dy}{dx} - \frac{dy}{dx} = 1 - y^{3}$$

That's fine. But on the next line you tried to divide through by x, getting:

$$3y^{2}\frac {dy}{dx}-\frac{dy}{dx} = \frac {1-y^{3}} {x}$$

That's wrong. The right side got divided by x, and the first term on the left got divided by x. The second one didn't. If you are going to divide by x, you have to divide EVERYTHING by x.

 

[Asphyxiated]

ok so can I do it this way and make:

$$x * 3y^{2}\frac{dy}{dx} - \frac{dy}{dx} = 1 - y^{3}$$

into

$$3y^{2}x\frac{dy}{dx} - \frac {dy}{dx} = 1-y^{3}$$

then

$$\frac {dy}{dx}(3y^{2}x-1) = 1-y^3$$

$$\frac {dy}{dx} = \frac {1-y^{3}} {3y^{2}x-1}$$

 

[Dick]

ok so can I do it this way and make:

$$x * 3y^{2}\frac{dy}{dx} - \frac{dy}{dx} = 1 - y^{3}$$

into

$$3y^{2}x\frac{dy}{dx} - \frac {dy}{dx} = 1-y^{3}$$

then

$$\frac {dy}{dx}(3y^{2}x-1) = 1-y^3$$

$$\frac {dy}{dx} = \frac {1-y^{3}} {3y^{2}x-1}$$

Now that looks right.

 

[Asphyxiated]

Alright cool, that division step was the only thing that I questioned myself about and I was right that I was wrong... haha anyway, just to make sure, the end result is as simple as possible right? I don't see anything to be done but I wanted to make sure.

Anyway thanks once again!

 

[Dick]

It's simple enough. Just stop making those dumb mistakes, ok?

 

[Asphyxiated]

yeah man, I'll try

 

[Asphyxiated]

So let's try this again, as long as it's ok with you:

$$y = \sqrt{xy+1}$$

which is the same as:

$$y = (xy+1)^{1/2}$$

so

$$\frac {d(y)}{dx} = \frac {d((xy+1)^{1/2})}{dx}$$

$$\frac {dy}{dx} = \frac {1}{2} (xy+1)^{-1/2} * (x\frac{dy}{dx} + y)$$

which is the same as:

$$\frac{dy}{dx} = \frac{1}{2} \frac{1}{\sqrt{xy+1}} * (x\frac{dy}{dx}+y)$$

more simple:

$$\frac {dy}{dx}= \frac {1} {2\sqrt{xy+1}} * (x\frac{dy}{dx}+y)$$

$$\frac {dy}{dx}= \frac{1}{2\sqrt{xy+1}}x\frac{dy}{dx}+\frac {1}{2\sqrt{xy+1}}y$$

$$\frac {dy}{dx} - \frac {1}{2\sqrt{xy+1}}y = \frac{1}{2\sqrt{xy+1}}x\frac{dy}{dx}$$

$$- \frac {1}{2\sqrt{xy+1}}y = \frac{1}{2\sqrt{xy+1}}x\frac{dy}{dx} - \frac {dy}{dx}$$

$$- \frac {1}{2\sqrt{xy+1}}y =\frac{dy}{dx} (\frac {1}{2\sqrt{xy+1}}x - 1)$$

$$\frac {-\frac{1}{2\sqrt{xy+1}}y} {\frac {1} {2\sqrt{xy+1}}x -1} = \frac {dy}{dx}$$

which is to say:

$$\frac {-\frac{y}{2\sqrt{xy+1}}} {\frac {x} {2\sqrt{xy+1}} -1} = \frac {dy}{dx}$$

then:

$$\frac {-\frac{y}{2\sqrt{xy+1}}} {\frac {x-2\sqrt{xy+1}} {2\sqrt{xy+1}}} = \frac {dy}{dx}$$


$$-\frac {y} {2\sqrt{xy+1}} * \frac {2\sqrt{xy+1}} {x-2\sqrt{xy+1}} = \frac {dy}{dx}$$

$$-\frac {(y)(2\sqrt{xy+1})} {(2\sqrt{xy+1})(x-2\sqrt{xy+1})} = \frac {dy}{dx}$$

$$-\frac {y} {x-2\sqrt{xy+1}} = \frac {dy}{dx}$$

so in summation:

$$\frac {dy}{dx} = -\frac {y} {x-2\sqrt{xy+1}}$$

I'm feeling confident this time, are you?

 

[snipez90]

That looks correct.

 

[Asphyxiated]

hoorah

 

[Asphyxiated]

I am hoping my addition of binomials is correct:

$$xy^{2} - 3x^{2}y +x =1$$

$$\frac {d(xy^{2}-3x^{2}y+x)}{dx} = \frac {d(1)}{dx}$$

$$(2yx\frac{dy}{dx}+y^{2}) - (3x^{2}\frac{dy}{dx}+6xy)+1 = 0$$

then I change the subtraction to addition and move the 1:

$$(2yx\frac{dy}{dx}+y^{2}) + (-3x^{2}\frac{dy}{dx}-6xy) = -1$$

$$2yx\frac{dy}{dx} - 3x^{2}\frac{dy}{dx} +y^{2}-6xy = -1$$

$$2yx\frac{dy}{dx} - 3x^{2}\frac{dy}{dx} +y^{2} = -1 + 6xy$$

$$2yx\frac{dy}{dx} - 3x^{2}\frac{dy}{dx} = -1 + 6xy - y^{2}$$

$$\frac{dy}{dx}(2yx - 3x^{2}) = -1 + 6xy - y^{2}$$

$$\frac{dy}{dx}= \frac {-1 + 6xy - y^{2}} {2yx - 3x^{2}}$$

 

[Dick]

I'm with you up to here:
$$\frac{dy}{dx}= \frac {-1 + 6xy - y^{2}} {2yx - 3x^{2}}$$

That's good. It's actually good work, then you went completely off the radar using some algebra rule you apparently just made up. Will you stop doing that?
This is getting to the point were I'm going to have to ask you not only to show your steps but to show the justification for each step. Otherwise, you are never going to learn to self-check these steps. You do fine for a while and then you just make a completely bongo move. Do you know that the manipulations you do in algebra are largely confined to i) adding or subtracting the same quantity from both sides an equation and ii) multiplying or dividing both sides of an equation by the same quantity? They aren't just like, randomly rearranging symbols in vaguely plausible ways.

How would you justify the step after that where you just bleeped out an xy from the numerator and denominator?

 

[Asphyxiated]

well I checked everything up to the point that I messed up, I thought that same values in the numerator and the denominator cancel. Since x*y = y*x and x=x and y=y. I definitely didn't just rearrange the symbols in vaguely plausible ways... but after checking the simplification (i will call it that) i see that I was wrong in that regard. Really the problem lies in me having to recall the correct way some step is done that I haven't used much recently, and vaguely remember from high school, I removed the two last steps since they are wrong

Not that it matters much since I evidently made up my own rule on this problem but did you look at the problem posted before this one? Pretty sure that one is 100% the first time around. I should just stop making assumptions when posting a problem. The last result I have on my board for the problem you just posted about is where you said that it was good. Usually I'll think that I see something else that can be done after I get it all written up and don't bother to check at that point... So, yeah

 

[Dick]

Sure. The main point is you can add/subtract to both sides of equation as in a=b+c -> a-c=(b+c)-c -> a-c=b or you can multiply/divide in a similar pattern. That's all. Ask yourself if that's REALLY what you are doing before you do anything. Cancelling the xy doesn't fall into either of those categories. Then maybe you won't have to check so much.

 

[Asphyxiated]

$$2xy^{-2}+x^{-2} =y$$

$$-4xy^{-3}\frac{dy}{dx} + 2y^{-2} - 2x^{-3} = \frac {dy}{dx}$$

$$-4xy^{-3}\frac{dy}{dx} + 2y^{-2} = \frac {dy}{dx} + 2x^{-3}$$

$$-4xy^{-3}\frac{dy}{dx} = \frac {dy}{dx} + 2x^{-3}- 2y^{-2}$$

$$-4xy^{-3}\frac{dy}{dx} -\frac {dy}{dx} = 2x^{-3}- 2y^{-2}$$

$$\frac{dy}{dx}(-4xy^{-3} -1) = 2x^{-3}- 2y^{-2}$$

$$\frac{dy}{dx} = \frac {2x^{-3}- 2y^{-2}} {-4xy^{-3} -1}$$