Question

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy.

x^3y^2 − x^4y + 4xy^3 = 0

Answer 1

@zepdrix

Answer 2

Ooo they're switching up the x and y's on you.
So sneaky! :O

Answer 3

I know! lol do I just do a normal product rule?

Answer 4

Yes, start with normal product rule.
But since we're taking the derivative `with respect to y`,
any `non-y` derivative we take will result in an extra little nugget.

So like when you take the derivative of a function of x, you should also get an x' attached to it.

Answer 5

so like 3x^2dx/dy?

Answer 6

yes good.

Answer 7

Primes are better though ^^ Less cumbersome lol.

Answer 8

3x^2' * y^2 + 2y * x^3 - 4x^3' * y + x^4 * 1 +4' * y^3 + 3y^2 * 4x

Answer 9

Ok a few boo boos to fix up real quick.

Answer 10

The prime goes on a separate x, not on the one you differentiated.\[\Large\bf\sf 3x^2x'y^2+2x^3y-4x^3x'y\color{red}{+}x^4(1)+4x'y^3+12xy^2=0\]

Answer 11

We also need to fix this red part.
See how there is a negative in front of your second term?
That negative gets `distributed` to each term in your product rule setup.

Answer 12

\[\Large\bf\sf \frac{dx}{dy}\quad=\quad x'\]

Answer 13

wait so
3x^2x′y^2 +2x^3 y−4x^3x′y - x^4 (1) - 4x′y^3 -12xy^2 =0 ???

Answer 14

Mmmmm yah that looks good! :D

Answer 15

:) yay!!

Answer 16

And now we need to solve for x'.
Any confusion so far? :U

Answer 17

a little but you explain well so I am ok lol

Answer 18

alright how do I solve for x'
combine like terms?

Answer 19

You won't be able to combine like terms since there is a big mess of x's and y's.
We'll instead need to `factor out` an x' from each term that contains an x'.