Some Implicit Differentiation (1 Viewer)

[Kirsti]

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]

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

(x+y)³ = (x^2 + y)²

x³ + 3x²y + 3xy² + y³ = x⁴ + 2x²y + y²

x⁴ - x³ - x²y - 3xy² +y² - y³ = 0

d/dx x⁴ = 4x³
d/dx -x³ = -3x²
d/dx -x²y = -2x.dy/dx
d/dx -3xy² = -6y.dy/dx
d/dx y² = 2y.dy/dx (I'm not 100% sure about those last 2)
d/dx -y³ = -3y².dy/dx

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

 

[Kirsti]

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

 

[Kirsti]

*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]

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]

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]

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

 

[Kirsti]

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]

(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]

(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]

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

Now lets do it the ^6 way:

(x + y)³ = (x² + y)²
3(1 + dy/dx)(x + y)² = 2(2x + dy/dx)(x² + y)
3(x + y)² - 4x(x² + y) = dy/dx(2)(x² + y) - dy/dx(3)(x + y)²

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

 

[Kirsti]

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

 

[KeypadSDM]

No problem. But that question is a bitch.

 

[maniacguy]

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]

 

[Affinity]

lol at maniac guy's signature

 

[KeypadSDM]

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.