Question

help please number 9



http://assets.openstudy.com/updates/attachments/575782a1e4b04b1a71f7aa00-marcelie-1465352875714-math185review.pdf

Answer 1

@Photon336

Answer 2

implicit differentiation

Answer 3

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

Answer 4

\[\tan(x-y) = \tan(u)*(du/dx)\]

let's start with the left hand side first

\[\sec^{2}(x-y)*(-\frac{ dy }{ dx }) = (y*(1+x^{2})^{\frac{ 1 }{ 2 }})~\]

Answer 5

wait wouldnt it be on the right side

sec ^2(x-y) (1-1dy/dx ) ?

Answer 6

yeah I totally missed that

Answer 7

\[\sec^{2}(x-y)(1-\frac{ dy }{ dx })\]

Answer 8

yeah nice work @marcelie

Answer 9

no no im just making sure that im not mistaken bc math could be complicated @agent0smith

Answer 10

@Photon336 what are you doing? where did the square root come from?

Answer 11

LOL OMG

Answer 12

D:

Answer 13

\[y(1+x^{2})^{-1} \]


u = (1+x^2)^-1 du\[\frac{ -2xy }{ (1+x^{2})^{2} }+\frac{ 1 }{ (1+x^{2}) }*\frac{ dy }{ dx }\] -
(1+x^2)^-2* -2x
v = y dv = dy/dx

Answer 14

\[u = (1+x^{2})^{-1} du = -(1+x^{2})^{-2}*2x \]

\[v = y ; dv = \frac{ dy }{ dx }\]

Answer 15

after this i'm taking a break

Answer 16

lool hmm

Answer 17

@Photon336 i feel like the quotient rule makes things a little easier here. \[\large (\frac{ y }{ 1+x^2 })'=\frac{ y' (1+x^2)-2xy }{ (1+x^2)^2 }\]

Answer 18

okay this where i got too and idk what to do next one sec

Answer 19

\[\sec^2 (x-y) (1-\frac{ dy }{ dx }) = (1+x^2)\frac{ dy }{ dx }-2xy \]

Answer 20

From there it's just algebra to get dy/dx alone (i used y' just cos it's a fraction in a fraction otherwise.

Your denominator is missing on the right @marcelie

Answer 21

so then the right side would look like this \[\sec^2 (x-y) (1-\frac{ dy }{ dx }) = (1+x^2)\frac{ dy }{ dx }-2xy / (1+x^2)^2\]

Answer 22

wait ill do the right side one sec

Answer 23

\[\sec^2 (x-y) (1-\frac{ dy }{ dx }) = \sec^2(x-y) - \sec^2(x-y) \frac{ dy }{ dx }\]

Answer 24

I didn't see that yeah that's much less work

Answer 25

@marcelie what happened on the right side?

Answer 26

I see what you mean now @agent0smith had to re-arrange this to get

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

Answer 27

wouldn't the right side look like that ? @agent0smith

Answer 28

Oh nvm i see what you were doing earlier marcelie. You were simplifying the LEFT side, but it looked like you'd massively changed the right side

Answer 29

yeh thats correct @Photon336 but should that fraction split ?

Answer 30

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

Answer 31

how do we go from here?

Answer 32

multiply (x+x^2)^2 to left side and right side ?

Answer 33

yeah then we would have to find a way to get y' by itself

Answer 34

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

Answer 35

Yeah i'd do that ^ then just distribute things, and get all the y' on the same side

Answer 36

wait where did you get (1-y') ?

Answer 37

(x-y) => 1-y'

Answer 38

1-dy/dx from earlier

Answer 39

\[\sec^{2}(x-y)-y'(\sec^{2}(x-y))*(1+x^{2}) = (1+x^{2})-(1+x^{2})y'\]

Answer 40

y' is often more convenient to write than dy/dx

Answer 41

okay so was there distribution ?

Answer 42

yeah I think distribution is what we can do to get y' by itself

Answer 43

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

Answer 44

hmm okay one sec

Answer 45

yeah show us what you get

Answer 46

what are you guys getting for the final answer?

Answer 47

not sure ive been trying to solve it but i got stuck many times D:

Answer 48

my answer looks nothing like the one at the end.

Answer 49

hmm

Answer 50

Oh sorry I kinda checked out. I think all was correct here: (i ain't showin all work)
\[\large \sec^{2}(x-y)(1-y')*(1+x^{2})^{2} = (1+x^{2})y'-2xy\]
\[\large -y'(1+x^{2})^{2} \sec^{2}(x-y)-y'(1+x^{2})= -2xy -(1+x^{2})^{2} \sec^{2}(x-y)\]
\[\large y'[(1+x^{2})^{2} \sec^{2}(x-y)+(1+x^{2})]= 2xy +(1+x^{2})^{2} \sec^{2}(x-y)\]
(mult both sides by -1)
\[\large y'= \frac{ 2xy +(1+x^{2})^{2} \sec^{2}(x-y) }{ (1+x^{2})^{2} \sec^{2}(x-y)+(1+x^{2})}\]
lastly factor out a (1+x^2) in denom:
\[\large y'= \frac{ 2xy +(1+x^{2})^{2} \sec^{2}(x-y) }{ (1+x^2)[(1+x^{2}) \sec^{2}(x-y)+1]}\]

Answer 51

Which is equivalent to the answer key.

Answer 52

okay the distribution part confusing me D:

Answer 53

on the right side

Answer 54

on the right...? this is the left\[\large \sec^{2}(x-y)(1-y')*(1+x^{2})^{2} = (1+x^{2})y'-2xy\]
change the order a lil bit \[\large \sec^{2}(x-y)(1+x^{2})^{2}(1-y') = (1+x^{2})y'-2xy\]
distribute that big ugly thing \[\large \sec^{2}(x-y)(1+x^{2})^{2}-\sec^{2}(x-y)(1+x^{2})^{2}y' =\]
then just moving terms around to get y' on the same side

Answer 55

oh yes sorry lmao im kind of sleepy lool

Answer 56

You'll prob want to do it on paper yourself. Just be very careful to keep track of the large terms

Answer 57

oh okay question for the bottom denominator where did the +1 come from ?

Answer 58

oh wait nvm lol

Answer 59

okay got it . ty <3 someone medal agent for me :)

Answer 60

You're welcome.