Question

Implicit Differentiation Problem... Help me?

Answer 1

Find dy/dx - assume a,b, and c are constants.


\[\arctan(x^2y) = xy^2\]

Answer 2

where are the a, b, c?

Answer 3

@satellite73 this is just from my book (no constants in this problem)

Answer 4

this is pretty much going to suck, but i bet we can do it

Answer 5

derivative of arcangent?

Answer 6

\[1/y^2+1 * dy/dx\]

Answer 7

and yeah it gets messy lol

Answer 8

if i recall it is \[\frac{1}{x^2+1}\]so start with \[\frac{1}{(x^2y)^2+1}\], then multiply by the derivative of \(x^2y\)

Answer 9

did i lose you at that step?

Answer 10

oh sorry no i got ya.

Answer 11

i actually did a lot of work for this problem but can't simplify it to look like the answer in the book

Answer 12

do you get that the derivative of \(x^2y\) is \(x^2y'+2xy\)?

Answer 13

yep

Answer 14

and on the right you get \[y^2+2xyy'\]if i am not mistaken

Answer 15

hmm... yeah i got that

Answer 16

then a raft of really ugly algebra to come

Answer 17

\[\frac{1}{(x^2y)^2+1}(x^2y'+2xy)=y^2+2xyy'\] solve for \(y'\)

Answer 18

so for i got...

\[(2xy+y^2y')/(x^4y^4+1)= (y^2 + 2yxy')\]

Answer 19

i would not mess with the denominator

Answer 20

okay haha

Answer 21

which is wrong as you wrote it, exponents messed up

Answer 22

oh yeah sorry i wrote it right in my notebook XD

Answer 23

y^2

Answer 24

x^2 idk so what next?

Answer 25

\[\frac{1}{(x^2y)^2+1}(x^2y'+2xy)=y^2+2xyy'\] then maybe \[\frac{x^2y'}{(x^2y)^2+1}+\frac{2xy}{(x^2y)^2+1}=y^2+2xyy'\]

Answer 26

put the stuff with \(y'\) on one side, all else on the other

Answer 27

really could this be any uglier?

Answer 28

i know lol -.-

Answer 29

all i did was distribute

Answer 30

\[\frac{x^2y'}{(x^2y)^2+1}+\frac{2xy}{(x^2y)^2+1}=y^2+2xyy'\]
\[\frac{x^2y'}{(x^2y)^2+1}-2xyy'=y^2-\frac{2xy}{(x^2y)^2+1}\]

Answer 31

What does the answer in the book look like?

Answer 32

not so bad if you copy and paste
now factor out the \(y'\) on the left

Answer 33

\[\frac{x^2y'}{(x^2y)^2+1}-2xyy'=y^2-\frac{2xy}{(x^2y)^2+1}\]
\[\left(\frac{x^2}{(x^2y)^2+1}-2xy\right)y'=y^2-\frac{2xy}{(x^2y)^2+1}\]

Answer 34

@skullpatrol
\[\frac{ y^2 + x^4y^4 - 2xy }{ x^2 - 2xy - 2x^5y^3 }\]

Answer 35

:O

Answer 36

which i have no idea how they simplified it to!

Answer 37

algebra

Answer 38

lol yeah XD

Answer 39

i wouldn't give it another thought

Answer 40

\[\frac{x (-2 x^4 y^3 + x - 2 y)}{(x^4 y^2 + 1)}y'=y^2-\frac{2xy}{(x^2y)^2+1}\]

Answer 41

and so on
really you are done
i would just multiply by the reciprocal and say i was finished

Answer 42

i'm with ya. thank you lol XD

Answer 43

yw
really i think you did all the work, the algebra leave to wolfram

Answer 44

Perhaps try a few test numbers to see if they are the same

Answer 45

hmm... yeah i wonder if you could just multiply everything by a (x^2y)^2+1 to simplify...? XD anyways i know theyre the same the rest is just algebra

Answer 46

\[\frac{x (-2 x^4 y^3 + x - 2 y)}{(x^4 y^2 + 1)}y'=y^2-\frac{2xy}{(x^2y)^2+1}\]
\[\frac{x (-2 x^4 y^3 + x - 2 y)}{(x^4 y^2 + 1)}y'=\frac{y (x^4 y^3 - 2 x + y)}{(x^4 y^2 + 1)}\]

Answer 47

what fun...

Answer 48

actually one more easy step

Answer 49

@satellite73 yeah i think you got it

Answer 50

multiply by the reciprocal, the \(x^2y^2+1\) will cancel

Answer 51

whew
got a harder one?

Answer 52

XD

Answer 53

@satellite73 thank you!

Answer 54

your welcome