Question

Another implicit differential question. Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy.
(x^3)(y^2)-(x^3)y+2x(y^3)=0

Answer 1

here is what i did:

Answer 2

\[3x^2y^2\frac{ dx }{ dy }-(3x^2y \frac{ dx }{ dy }+x^3)+(2(y^3\frac{ dx }{ dy }+3xy^2))\]
\[3x^2y^2\frac{ dx }{ dy }-3x^2y \frac{ dx }{ dy }+2y^3\frac{ dx }{ dy }=x^3-2y^3-6xy^2\]
\[\frac{ dx }{ dy }=\frac{ (x^3-2y^3-6x^2y) }{ 2y^3 }\]

Answer 3

but apparently this is wrong, can anyone else find my mistake

Answer 4

Ooo I see :O
You forgot to product rule on the first term.

Answer 5

maybe? hold on checking again XD

Answer 6

my bad just forgot to put in this, but on paper i did put it

Answer 7

\[\large\rm x^3y^2-x^3y+2xy^3=0\]It's somewhere in the derivative of the last term.
Hmm.

Answer 8

ooh i know what i messed up in

Answer 9

never mind, its still wrong lol

Answer 10

This is another sneaky trick you could have applied to the left side,
\[\large\rm x^3(y^2-y)+2xy^3=0\]which saves you from doing the product rule one extra time.

Just a thought :)

Answer 11

I don't understand what you did, can you elaborate?

Answer 12

I factored an x^3 out of the first two terms: \(\large\rm \color{orangered}{x^3}y^2-\color{orangered}{x^3}y=\color{orangered}{x^3}(y^2-y)\)
Then you do your product rule a little bit differently for that first big term,
\[\large\rm \color{royalblue}{(x^3)'}(y^2-y)+x^3\color{royalblue}{(y^2-y)'}+\color{royalblue}{(2x)'}y^3+2x\color{royalblue}{(y^3)'}=0\]

Answer 13

oh ok

Answer 14

It allows you to do product rule twice, instead of three times.
Not a big deal though.

Answer 15

And I recommend using \(\large\rm x'\) instead of \(\large\rm \frac{dx}{dy}\).
It makes the math so much easier to read.
But that's just my own preference I guess :)

Answer 16

I can't seem to figure out where this is coming from:\[\large\rm 3x^2y^2\frac{ dx }{ dy }-3x^2y \frac{ dx }{ dy }+2y^3\frac{ dx }{ dy }=x^3\color{red}{-2y^3}-6xy^2\]The one in red.
Hmm

Answer 17

what about it?

Answer 18

thats from the first term

Answer 19

oh wait hmm

Answer 20

ooh thats suppose to 2yx^3

Answer 21

I must have messed when I was collecting like terms

Answer 22

(x^3)(y^2)
-(x^3)y
+2x(y^3)
=0


(x^3)'(y^2)+(x^3)(y^2)'
-(x^3)'y-(x^3)y'
+2x'(y^3)+2x(y^3)'
=0'

Answer 23

Can you guys check if this is right?
\[\frac{ x^3-2yx^3-6xy^2 }{ 3x^2 y^2-3x^2y+2y^3}\]

Answer 24

this is what I got for my final answer after fixing some mistakes

Answer 25

Ooo that looks better! :)

Answer 26

(3x^2x')(y^2)+(x^3)(2yy')
-(3x^2x')y-(x^3)y'
+2x'(y^3)+2x(3y^2 y')
=0

y' = dy/dy = 1


(3x^2x')(y^2)+(x^3)(2y)
-(3x^2x')y-(x^3)
+2x'(y^3)+2x(3y^2)
=0

x'*
(3x^2)(y^2)
-(3x^2)y
+2(y^3)
= -2x(3y^2)+(x^3)-(x^3)(2y)

etc ...

Answer 27

awsome thanks guys