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Find dy/dx by implicit differentiation

  • Thread starter momogiri
  • Start date
  • #26
52
0
Ok, so what I did was I got 0.5(xy)^-0.5, then into 0.5(xy'+y)^-0.5
Then I figured out the other side, in which the 2 is gone, and d/dx of x^2y is x^2y' + 2xy

Then.. I'm trying to solve for dy/dx, so..
[tex]\frac{1}{\sqrt{0.5(xy'+y)}}=x^{2}y' + 2xy[/tex]
Which becomes
[tex]\frac{1}{\sqrt{0.5(xy'+y)}} - x^{2}y' = 2xy[/tex]
And then... I'm stuck :(
 
  • #27
492
1
[tex](\sqrt{xy})'\neq\frac{1}{\sqrt{0.5(xy'+y)}}[/tex]

you had it partially right with:
[tex]\frac{1}{2\sqrt{xy}}[/tex]

you just need to continue using the chain rule since you have (f(g(x)))'=f'(g(x))g'(x) you did f'(g(x)) now you need to multiply by g'(x). where g(x)=xy
 
  • #28
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So the part I'm missing is the g'(x) of [tex]g'(x)*\frac{1}{\sqrt{0.5(xy'+y)}}[/tex]?
 
  • #29
492
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no you're missing g'(x) of [tex]g'(x)*\frac{1}{2\sqrt{xy}}[/tex]
 
  • #30
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Oh...
Where did the [tex]\frac{1}{2\sqrt{xy}}[/tex] come from? XD

And g'(x) = (xy)' ?
I'll have to use the product rule then, right?
 
  • #31
492
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yes, and that fraction is the f'(g(x)) of the function [tex]\sqrt{xy}[/tex]

since [tex](\sqrt{u})'=\frac{u'}{2\sqrt{u}}[/tex]
 
  • #32
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Oh ok, so [tex]\frac{1}{2\sqrt{xy}}*(xy'+y)=x^{2}y' + 2xy[/tex]?
 
  • #33
492
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depends, was the right hand side originally [tex]2+2xy[/tex] or [tex]2+x^2y[/tex]?

if it was yx^2 then yes it's right, if it wasn't then no
 
  • #34
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Yep, it was originally [tex]2+x^2y[/tex]

So [tex]\frac{1}{2\sqrt{xy}}*(xy'+y)=x^{2}y' + 2xy[/tex]
becomes
[tex]\frac{1}{2\sqrt{xy}}xy'+\frac{1}{2\sqrt{xy}}y=x^{2}y' + 2xy[/tex]
which I shuffle..
[tex]\frac{1}{2\sqrt{xy}}xy' - x^{2}y' = 2xy - \frac{1}{2\sqrt{xy}}y[/tex]
and stuff..
[tex]y'(\frac{1}{2\sqrt{xy}}x - x^{2}) = 2xy - \frac{1}{2\sqrt{xy}}y[/tex]
thus
[tex]y' = \frac{2xy - \frac{1}{2\sqrt{xy}}y}{\frac{1}{2\sqrt{xy}}x - x^{2}}[/tex]?
 
  • #35
492
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correct, you could also reduce it a bit more by getting everything with a common denominator of 2sqrtxy to cancel that out.
 
  • #36
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You mean by time-sing 2xy and x^2 by 2sqrtxy?
 
  • #37
492
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yes and combining the fractions on the top and the bottom you can cancel out 2sqrt(xy) and have just 1 denominator/numerator.
 
  • #38
52
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Hmm.. I'm not sure if I did it correctly though, so here it is:
[tex]\frac{2xy - \frac{1}{2\sqrt{xy}}y}{\frac{1}{2\sqrt{xy}}x - x^{2}}[/tex]
into
[tex]\frac{\frac{2xy(2\sqrt{xy})}{(2\sqrt{xy})} - \frac{y}{2\sqrt{xy}}}{\frac{x}{2\sqrt{xy}} - \frac{x^{2}(2\sqrt{xy})}{(2\sqrt{xy})}}[/tex]
and into
[tex]\frac{\frac{2xy(2\sqrt{xy}) - y}{2\sqrt{xy}}}{\frac{x - x^{2}(2\sqrt{xy})}{2\sqrt{xy}}}[/tex]
which becomes
[tex]\frac{2xy(2\sqrt{xy}) - y}{x - x^{2}(2\sqrt{xy})}[/tex]
and..
[tex]\frac{4xy(\sqrt{xy}) - y}{x - x^{2}(2\sqrt{xy})}[/tex]?
I donno if I can simplify it more XP
 
  • #39
492
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yes that's about as simplified as you can go
 
  • #40
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Man, that was a pain.. but.. I got stuck again :P Surprise surprise..

Question:
Find dy/dx by implicit differentiation.
[tex]e^{x/y} = 2x - y[/tex]


Attempted:
Isn't [tex](d/dx)e^{x/y} = e^{x/y}[/tex]?
Well, I went ahead anyway, and figured out
[tex](d/dx)2x = 2[/tex] and [tex](d/dx)y = y'[/tex]
Then I um.. rearranged it so it's
[tex]y' = 2 - e^{x/y}[/tex]
Which is wrong. lol

I'm assuming that [tex](d/dx)e^{x/y} \neq e^{x/y}[/tex] :P Then how do I find it?
 
  • #41
492
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[tex](e^u)'=u'e^u[/tex]

so you need to use the quotient rule on x/y and multiply that to e^(x/y)
 
  • #42
52
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Damn, lol
I'll try it XD

Dang, I thought I had it right, but I got it wrong :P
Here's what I did..
Quotient rule = [tex]\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}[/tex]
So I plugged in the numbers as usual..
[tex]\frac{(y)(x')-(x)(y')}{(y)^{2}}[/tex] which got me [tex]\frac{(y)-(xy')}{y^{2}}[/tex]
Therefore the whole thing becomes
[tex]\frac{y-xy'}{y^{2}}e^{\frac{x}{y}} = 2 - y'[/tex]
So I got rid of y^2
[tex]y-e^{\frac{x}{y}}xy' = 2y^{2} - y^{2}y'[/tex]
then shuffled
[tex]y-2y^{2} = e^{\frac{x}{y}}xy' - y^{2}y'[/tex]
Thus
[tex]y-2y^{2} = y'(e^{\frac{x}{y}}x - y^{2})[/tex]
Therefore
[tex]y' = \frac{y-2y^{2}}{e^{\frac{x}{y}}x - y^{2}}[/tex]...?

What did I do wrong this time :/
 
Last edited:
  • #43
492
1
when you got rid of y^2 you didn't keep the parenthesis:
(y-xy')e^(x/y) =/=y-e^(x/y)xy'

so you're just missing a e^(x/y) multiplied to y in the numerator
 
  • #44
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Eeek, I don't see it..
And by getting rid of y^2, I actually meant timesing it to the other side XD
 
  • #45
492
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[tex]\frac{(y-xy')e^{\frac{x}{y}}}{y^2}\neq y-e^{\frac{x}{y}}xy'[/tex]
and the y^2 being carried over to the other side.
it should be:

[tex]ye^{\frac{x}{y}}-xy'e^{\frac{x}{y}}[/tex]

and continuing from there you are only missing that e^(x/y) on your final answer
 
  • #46
52
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Thanks for that, bob :D
I've got another question :/
Use implicit differentiation to find an equation of the tangent line to the curve at the point (0, -0.5).
[tex]x^{2} + y^{2} = (2x^{2} + 2y^{2} - x)^{2}[/tex]


Ok, so [tex]x^{2} + y^{2} = (2x^{2} + 2y^{2} - x)^{2}[/tex] becomes [tex]x^{2} + y^{2} = 2x^{4} + 2y^{4} - x^{2}[/tex]
Which I solved for the d/dx
[tex]2x + 2yy' = 8x^{3} + 4(2y)^{3}*2y' - 2x[/tex]
Shuffling..
[tex]2yy' - 4(2y)^{3}*2y' = 8x^{3} - 4x[/tex]
and again
[tex]y'(2y - 8(2y)^{3}) = 8x^{3} - 4x[/tex]
thus
[tex]y' = \frac{8x^{3} - 4x}{2y - 8(2y)^{3}}[/tex] which equals
[tex]y' = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}[/tex]

Ok. So that's my slope for the tangent line..
And my equation for tangent line is
[tex]y = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}x[/tex]
and at point (0,-0.5)
[tex]y + 0.5 = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}x[/tex]
Making it
[tex]y = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}x - 0.5[/tex]
So I've managed to solve up to here, and.. frankly, I don't know if I'm right XD
Oh, and this is an image they gave me
3-5-27.gif
 

Attachments

  • #47
492
1
no you have [tex](2x^2+2y^2-x)^2=(2x^2+2y^2-x)(2x^2+2y^2-x)[/tex]
so you have to multiply that out and then you can do the derivative.

or you can do it straight away [tex](u^n)'=nu^{n-1}u'[/tex]
where u=2x^2+2y^2-x

also in order to find the actual slope of the function at (0,-0.5) when you isolate y' plugin (0,-0.5) that way you'll get a number.
so your equation for the line shoudl be: y+0.5=mx
 
Last edited:
  • #48
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Oh XD Ok

(Off-topic) Hey Bob, will you be on later tonight?
I have to go eat dinner and travel back to my dorm :/ That'll take an hour or two, so...
It'd be great if you're still available to help me after XD

Edit: Daaang, just read your message XD
Oh well, I'll manage somehow :D I appreciate all your help :D
 
  • #49
492
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sorry but I'm sure other people will be able to help too,
just remember that when you need to do product/quotient rule and then do the chain rule if you have (y^n)' where n is different from 1.
 

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