find dy/dx of
X^3+y^3-4x^2y=0
given y is a differentiable function of x

Question

find dy/dx of 
X^3+y^3-4x^2y=0
given y is a differentiable function of x

anonymous · Answer

I understand this is implicit differentiation, but I have been having difficulty with the core concept. That is 4 times x^2 times y, for clarification.

anonymous · Answer

$$X ^{3}+ y^3-4x^2y=0$$

loser66 · Answer

cal1?

anonymous · Answer

Yes

anonymous · Answer

I feel likes its implicit, but I have difficulty with implicit differentiation

loser66 · Answer

what do you get so far?

anonymous · Answer

Frankly, it confused me enough I went back to attempt easier implict differentiation problems. I would need a walkthrough, at least for some part of it until it clicked.

anonymous · Answer

Am I correct in $$3x^2+2y^2-8x*y \prime=0$$

anonymous · Answer

that should be 3y^2, my error.

loser66 · Answer

$$3x^2+3y^2y'-(8x*y+4x^2y')=0$$

loser66 · Answer

the last term is 4x^2y consider as a product of 4x^2 and y, take derivative of product

loser66 · Answer

the middle term should be as I stated. (y^3)' = 3y^2 y'

anonymous · Answer

So, a mixture of implicit differentiation and the product rule?

loser66 · Answer

$$3x^2 +3y^2y'-8xy-4x^2y'=0\3y^2y'-4x^2y'=8xy-3x^2\y'(3y^2-4x^2)=8xy-3x^2\\frac{dy}{dx}=\frac{8xy-3x^2}{3y^2-4x^2}$$

loser66 · Answer

yes for your last question, since y is a function respect to x, when take derivative of it, it should be solved as I did

anonymous · Answer

Alright, this makes it very clear. Thank you very much for your help!

loser66 · Answer

yw