Some Implicit Differentiation (1 Viewer)

Kirsti

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I can't find my dy/dx. Can anybody help me look for it? (i feel a bit dumb, i've done the rest of the exercise just cant get this)


find dy/dx for

(x + y)^1/2 = (x^2 + y)^1/3


and


(x^2 + y^2)^-1/2 = 4


for the first i can get the numerator but my denominator is highly confused. the 2nd i'm just getting WRONG
 

Calculon

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(x + y)<sup>1/2</sup> = (x^2 + y)<sup>1/3</sup>

(x+y)<sup>3</sup> = (x^2 + y)<sup>2</sup>

x<sup>3</sup> + 3x<sup>2</sup>y + 3xy<sup>2</sup> + y<sup>3</sup> = x<sup>4</sup> + 2x<sup>2</sup>y + y<sup>2</sup>

x<sup>4</sup> - x<sup>3</sup> - x<sup>2</sup>y - 3xy<sup>2</sup> +y<sup>2</sup> - y<sup>3</sup> = 0

d/dx x<sup>4</sup> = 4x<sup>3</sup>
d/dx -x<sup>3</sup> = -3x<sup>2</sup>
d/dx -x<sup>2</sup>y = -2x.dy/dx
d/dx -3xy<sup>2</sup> = -6y.dy/dx
d/dx y<sup>2</sup> = 2y.dy/dx (I'm not 100% sure about those last 2)
d/dx -y<sup>3</sup> = -3y<sup>2</sup>.dy/dx

Then just add that up and make dy/dx the subject
 
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Kirsti

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lol, don't know how or why but i think i left out a 3 in the 4th line
i might try this again
 
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Kirsti

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*sigh* nope

I'm geting the same answer i was before

the answer they're giving me for it is


(4x - 3(( x + y )^1/2))/(3(( x + y )^1/2) - 2)

and i cant get it
could someone work it through fully for me?
 

Calculon

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try making both sides go ^6, then expand the brackets and take it from there. I don't know if the quotient rule works for implicit diff.
 

Kirsti

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I did but I can't get the answer they want

hey, something i just noticed in your working, wouldn't y.x^2 and 3x.y^2 use the product rule?
 

Calculon

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Yeah i should, now you're helping me :p, its just i find it cofusing doing calculations without paper.
And also the answer may be wrong in the back of the book
 

Kirsti

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I think I'll just assume that the book is wrong. Thanks for the help. At least I know I've got a good method.
 

KeypadSDM

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(x + y)^1/2 = (x^2 + y)^1/3

(1 + dy/dx) * 1/2 * (x + 1)^-1/2 = (2x + dy/dx) * 1/3 * (x^2 + y)^-2/3
3(1 + dy/dx)(x + 1)^-1/2 = 2(2x + dy/dx)(x^2 + y)^-2/3
3(x + 1)^-1/2 + 3(dy/dx)(x + 1)^-1/2 = 4x(x^2 + y)^-2/3 + 2dy/dx(x^2 + y)^-2/3
3(x + 1)^-1/2 - 4x(x^2 + y)^-2/3 = 2dy/dx(x^2 + y)^-2/3 - 3(dy/dx)(x + 1)^-1/2
dy/dx = [3(x + 1)^-1/2 - 4x(x^2 + y)^-2/3]/[2(x^2 + y)^-2/3 - 3(x + 1)^-1/2]

I suppose you could probably simplify that ... Maybe
 

KeypadSDM

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(x^2 + y^2)^-1/2 = 4

(2x + 2y * dy/dx) * (-1/2) * (x^2 + y^2)^-3/2 = 0
(2x + 2y * dy/dx)(x^2 + y^2)^-3/2 = 0

I just realised I didn't need this bit, but I'll leave it in just in case I'm wrong:
2x(x^2 + y^2)^-3/2 = -(2y * dy/dx)(x^2 + y^2)^-3/2
- dy/dx = [2x(x^2 + y^2)^-3/2]/[2y(x^2 + y^2)^-3/2]


(2x + 2y * dy/dx)(x^2 + y^2)^-3/2 = 0
2x + 2y * dy/dx = 0
dy/dx = -x/y
 

KeypadSDM

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(x + y)^1/2 = (x^2 + y)^1/3

Now lets do it the ^6 way:

(x + y)<sup>3</sup> = (x<sup>2</sup> + y)<sup>2</sup>
3(1 + dy/dx)(x + y)<sup>2</sup> = 2(2x + dy/dx)(x<sup>2</sup> + y)
3(x + y)<sup>2</sup> - 4x(x<sup>2</sup> + y) = dy/dx(2)(x<sup>2</sup> + y) - dy/dx(3)(x + y)<sup>2</sup>

I can't factorize that, screw you guys, I'm going home.
 

Kirsti

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thankyou keypad

i cant work out what i did wrong yet but ill find it soon
(methinks i have a wrong sign or two)

thanks for showing that it can work though
 

maniacguy

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Kirsti, the way to get their answer is to use the original relation. (The book IS right, btw - it needs some manipulation to get there though!)

The solution:
(x+y)^(1/2) = (x^2 + y)^(1/3)

Differentiating wrt x:
(1+y')*1/2*(x+y)^(-1/2) = (2x+y')*1/3*(x^2+y)^(-2/3)

Now, (x^2+y)^(-2/3)
= [(x^2+y)^(1/3)]^(-2)
= [(x+y)^(1/2)]^(-2) (from our original relation)
= (x+y)^(-1)

[Keypad, this is where your first solution broke down. That substitution makes your answer equivalent to the right one. Well, if you exclude the 'x+1' that you turned 'x+y' into in the second step!]

So we have:
(1+y')*1/2*(x+y)^(-1/2) = (2x+y')*1/3*(x+y)^(-1)

Multiplying through by (x+y):
(1+y')*1/2*(x+y)^(1/2) = (2x+y')*1/3

Multiplying by 6 to get rid of the fractions:
3(1+y')*(x+y)^(1/2) = 2(2x+y')

(3+3y')*(x+y)^(1/2) = 4x+2y'

Now taking the y' terms over to one side:
y' * [ 3*(x+y)^(1/2) - 2 ] = 4x - 3*(x+y)^(1/2)

Dividing by [ 3*(x+y)^(1/2) - 2 ] then gets the desired result:
y' = [4x - 3*(x+y)^(1/2)]/[3*(x+y)^(1/2) - 2]
 

KeypadSDM

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Originally posted by maniacguy
[Keypad, this is where your first solution broke down. That substitution makes your answer equivalent to the right one. Well, if you exclude the 'x+1' that you turned 'x+y' into in the second step!]
Ahh, yes. Substituting back into the original equation.

I never did much implicit differentiation, it doesn't seem to be a huge thing in the course.
 

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