What is the Implicit Differentiation of the Equation x+y=1+x^3y^2?

[jkeatin]

Homework Statement

dy/dx: square root x+y= 1+x^3y^2

Homework Equations

chain rule
implicit differentiation

The Attempt at a Solution

1/2 x+y -1/2 =2x^2y^3 *y'

 

[rocomath]

\sqrt{x+y}=1+x^3y^2

Be clear! Make use of parenthesis. Right?

 

[jkeatin]

yeah, my bad

 

[jkeatin]

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

 

[jkeatin]

am i going in the right direction?

 

[rock.freak667]

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

To differentiate the LHS w.r.t x
1/2(x+y)^-1/2 is correct but you'll need to multiply it by the differential of (x+y) i.e. what is in the bracket.

For the RHS : 1+x^3y^2 use the product law for x^3y^2

 

[jkeatin]

ok
1/2(x+y)^-1/2 + 1/2(x+y)^-1/2 (y)'= 3x^2y^2 +2y (y)' (x^3)

 

[jkeatin]

is that right?

 

[Defennder]

Yes it is.

 

[jkeatin]

do i need to simplify anymore?

 

[Defennder]

Are you required to?

 

[jkeatin]

I need to find y'

 

[jkeatin]

is this the answer?
y'= 3x^2-1/2(x+y)^-1/2 over [1/2(x+y)^-1/2] - 2yx^3

 

[rocomath]

Yes it is.

Defennder confirmed your "Calculus steps" I'm sure you can handle the rest.