Implicit Differentiation: Finding dy/dx for a Given Equation

[Grogerian]

Homework Statement

Implicit Differentiation:

I Was given the equation find dy/dx:
(3x³y² + 7x)
(x²y³ + 3xy)-

The Attempt at a Solution

Ok, i know i have to use the product rule on top, and on bottom and the quotient rule for the fraction so... if i set

s = 3x³
t = y²
u = (3x³y² + 7x)
w = x²
n = y³
z = 3x
q = y
v = (x²y³ + 3xy)

ds = 9x²
dt = 2y(dy/dx)
du = (s(dt) + t(ds)) + 7)
dw = 2x
dn = 3y²(dy/dx)
dz = 3
dq = 1(dy/dx)
dv = (w(dn) + n(dw)) + (z(dq) + q(dz))

Quotient rule: v(du) - u(dv)/(v²)

ok, I'm just making sure i didn't make a mistake.

so now i do:

(x²y³ + 3xy)((3x³(2y(dy/dx)) + y²(9x²)) + 7)) - (3x³y² + 7x)((x²(3y²(dy/dx)) + y³(2x)) + ((3x(1(dy/dx)) + 3y)
(x²y³ + 3xy)²


I had a tutor help me out in Implicit Diff. but i still don't really remember this stuff unless i look at it, and i almost always get different from the professor, is the above correct ( without factoring?) it looks a lot easier on my paper lol. and I'm supposed to be ready for this kind of magnitude questions on my exam :S i also don't know how i am going to be able to get dy/dx out of my equation, any tips there?

 

[myk127]

you should group all the terms that has dy/dx in it on the other side. then factor out dy/dx.

 

[Grogerian]

ok, but once i have (dy/dx)*(long equation) i can't just subtract dy/dx to the other side?

 

[myk127]

ok, but once i have (dy/dx)*(long equation) i can't just subtract dy/dx to the other side?

you won't subtract it. you just have to divide the other side by the long equation you said.

 

[Grogerian]

well then, i feel stupid lol i even knew that ^.^ i differentiate the other side as well, which was = 7x -> 7