Question

find dy/dx if xcos(y) - y^2 = 4 - x^2

Answer 1

i got - [2x] / [ xsin(y) - cos(y) ]

Answer 2

implicit differentiation:

cos(y) + x* -sin(y) * dy/dx - 2y* dy/dx = -2x
2y* dy/dx - sin(y) x dy/dx = -2x - cos(y)
dy/dx = (-2x - cos(y) ) / (2y - x sin(y) )

Answer 3

hmm - i'm just checking my work....

Answer 4

i think i'm right - i'll check it out on wolfram

Answer 5

its 2x + cos (y)
----------
x sin(y) - 2y

which is same as my answer except they have multiplied top and bottom by -1 to make the whole thing positive

Answer 6

what's the first step.. i don't think i get it.

Answer 7

\[(xcos(y))'=(x)'\cos(y)+x(\cos(y))'=1 \cos(y)+x y' (-\sin(y))=\cos(y)-xy'\sin(y)\]
\[(y^2)'=2yy'\]
\[(4-x^2)=(4)'-(x^2)'=0-2x=-2x\]
So we have
\[\cos(y)-xy'\sin(y)-2yy'=-2x\]

Answer 8

what's deriv of cos(y)

Answer 9

-sin(y) * y' right?

Answer 10

differentiating x cos(y) is done using the product rule as myin has done above using the y' notation

cos)y) when differentiating w r t to x you regard y as a function of f
x an use chain rule

Answer 11

yes i know to use chain

Answer 12

yes -siny * y' is correct

Answer 13

ok- sorry

Answer 14

so i get -xy'sin(y) + cos(y) = -2x

Answer 15

is that correct so far?

Answer 16

no you forgot derivative of y^2 which is 2y* y'

Answer 17

ahhhhhhhhhhhhhh

Answer 18

-xy'sin(y) + cos(y) - 2y y' = -2x

Answer 19

i just saw an error - the second line should start with -2

Answer 20

so the denominator of the answer should read x sin(y) + 2y

Answer 21

phew!!!

Answer 22

-2y'xsin(y)-2y = -2x

Answer 23

good till there?

Answer 24

yea

Answer 25

-2y'xsin(y)-2y y' = -2x

Answer 26

so i get: - [2x] / [2xsin(y)+2y]

Answer 27

err

Answer 28

hmm - i'm going crazy here ( not your fault) - lets have another check

Answer 29

you've missed out the cos(y) alien

Answer 30

nah i forgot cos... lemme do it again

Answer 31

(-xy'sin(y))+(cos(y))-[2y*y'] = -2x

Answer 32

so far so good?

Answer 33

yea

Answer 34

ok so how do i combine the y'

Answer 35

take out the y' form 2 terms containing it
and subtract cos(y) from both sides

Answer 36

y'( -x sin(y) - 2y) = -2x - cos(y)

Answer 37

so finally i get: - [2x + cos(y)] / [xsin(y) + 2y]

Answer 38

no minus in the numerator

Answer 39

i factored out the minus from the whole thing

Answer 40

so before i did that it read: [-2x-cos(y)] / [xsin(y)+2y]

Answer 41

err

Answer 42

i mean that [-2x-cos(y)] / [-xsin(y)+2y]

Answer 43

ack

Answer 44

i mean this for real this time: [-2x-cos(y)] / [-xsin(y)-2y]

Answer 45

itd -2x - cos(y) / -xsin(y) - 2y
- yes
so when you multiply top and bottom by -1
all signs become positive

Answer 46

hurray

Answer 47

yayyy!!

Answer 48

i quit math now.

Answer 49

lol