Question

I've worked through this problem several times, but can't figure out how to get the answer the book has, or what I'm doing wrong. Can someone take a look at my process, and explain how to do it "right"? What am I messing up or forgetting? What should I be using?
The problem:
(x^4)(x+y) = (y^2)(3x-y); Find dy/dx using Implicit Differentiation.

Answer 1

This is how I've gone about finding dy/dx:

Answer 2

\[x^4(x+y) = y^2(3-y) \]
\[(4x^3)(x+y)+x^4(1+(dy/dx)=2y(dy/dx)(3x-y) + y^2(3(1)-(dy/dx))\]
\[4x^4 + 4x^3y+x^4+x^4(dy/dx) = 6xy(dy/dx) -2y^2(dy/dx)+3y^2-y^2(dy/dx)\]
\[5x^4+4x^3y+x^4(dy/dx) = 6xy(dy/dx) +3y^2 - 3y^2(dy/dx)\]

Answer 3

\[5x^4+4x^3y+x^4(dy/dx)=6xy(dy/dx)-3y^2(dy/dx)+3y^2\]
\[5x^4+4x^3y+-3y^2=6xy(dy/dx)-3y^2(dy/dx)-x^4(dy/dx)\]
\[5x^4+4x^3y-3y^2=dy/dx(6xy-3y^2-x^4)\]
\[dy/dx = \frac{5x^4+4x^3y-3y^2}{6xy-3y^2-x^4}\]

Answer 4

*Sorry for the last step looking funky – OS has been making fractions look off for a couple of weeks now at least... ?

Answer 5

No that's fine..

And everything looks perfect to me...what does the book have?

Answer 6

The book has:
\[\frac{3y^2-5x^4-4x^3y}{x^4+3y^2-6xy}\]

Answer 7

And I initially was like "Oh, I just multiply the top and bottom by -1, right?" But I must ask... why would I do that, and wouldn't that make it a NEGATIVE dy/dx then?

Answer 8

-1/-1=1

Answer 9

Yes thank you @freckles lol I couldnt type it quick enough

Answer 10

Exactly, so I'm like, super confused about how the signs are all switched, but the terms are the same...

Answer 11

Your answer is fine though. They probably did there adding and subtracting of things on both sides a bit differently than you. But yeah the answers are equal.

Answer 12

You guys sure? That was my gut reaction too when I read the book, but I don't want to get no credit, if it's some technical thing.

Answer 13

Yes I'm sure - 1/-1 is 1

Answer 14

Do it...multiply BOTH top and bottom by -1

it will change all the signs you want

And you're overthinking it...it will NOT equal -dy/dx

*Because as @freckles is saying...-1/-1 = 1

Answer 15

-dy/dx would be just changing the signs on either top or bottom...not both

Answer 16

But you're sure that both final answers are mathematically equivalent, that's what I'm pondering the truth of... ?
Alright. But why would it make sense to do this in this case? Where does it come from, if I worked my math out fine? If I had never looked at the back of the book, I would have assumed I had the right answer and would have done everything I could have, ostensibly being "done", right? How would I know to change the signs like this?

Answer 17

You wouldn't need to, they are mathematically equivalent...Admittedly, this does always throw me off as well *when a textbook has SUCH a similar answer in a different manner* however, in this case, they are the same

Answer 18

Thank you both, very much, I really appreciate the clarification :) It helps me learn sooo much better

Answer 19

No problem!

Answer 20

here is an example of why you are equivalently asking if -1/-1=1

say we have the equation:
-x+5=-2+2x
add x and add 2 on both sides giving us:
5+2=x+2x
5+2=x(1+2)
so x=(5+2)/(1+2) or just 7/3

now if I chose to subtract 2x and subtract 5 on both sides that would give us:
-x-2x=-2-5
x(-1-2)=-2-5
x=(-2-5)/(-1-2) which is the just -7/-3 which is just 7/3

So I was saying earlier they probably didn't take their answer and multiply -1/-1.
They probably just went about solving for dy/dx differently by choosing to add/subtract different terms than you on both sides.