# Homework Help: Implicit Differentiation

1. Oct 2, 2008

### Grogerian

1. The problem statement, all variables and given/known data
Implicit Differentiation:

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

2. Oct 2, 2008

### myk127

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

3. Oct 2, 2008

### Grogerian

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

4. Oct 2, 2008

### myk127

you wont subtract it. you just have to divide the other side by the long equation you said.

5. Oct 2, 2008

### Grogerian

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