Question

how do I take the implicit derivative of 4cosXsinY=1 ?

Answer 1

you'd have to take the d/dx of each side so you're going to have to use the product rule

Answer 2

4cos(x) = u
sin(y)= v

Answer 3

u'= -4sin(x)

Answer 4

Right, I got down to where I have 4(sinx-cos(dy/dx)) + (cosx)(-cos(dy/dx)) + (sinx)(siny)
and now I don't really know if that's right

Answer 5

Do you have any idea if or where I messed up?

Answer 6

what did you get before that

Answer 7

what did you get for the derivative of cos(y)

Answer 8

the answer is supposed to be Y' = tanXtanY

Answer 9

The first thing I did was to take the the derivitive of 4cosXsinY=1 so that the first thing I did was to come up with 4(sinx-cos(dy/dx)) + [(cosx)(siny)' + (cosx)'(siny)]

Answer 10

i think that is where you got off track

d/dx[sin(y)]=cos(y)dy/dx

Answer 11

Then the second thing I did was the equation I gave earlier: 4(sinx-cos(dy/dx)) + (cosx)(-cos(dy/dx)) + (sinx)(siny)

Answer 12

look at cos(y) as cos(u) i guess you could say so the derivative of this would be cos(u)u'

u=y
u'=dy/dx

Answer 13

Ok

Answer 14

i mean't sin(y) derivative = cos(u)u'

Answer 15

yeah

Answer 16

It has been a while since I had trig too so that isn't really helping too much either

Answer 17

so you should get

4cos(x)(cos(y)y-4sin(y)sin(x)=0
*add the 4sin(y)sin(x) to the left and factor out y' on the left

y'(4cos(x)cos(y))=4sin(y)sin(x)
*divide by 4cos(x)cos(y)
y'=tanYtanX

Answer 18

or TanXTanY same thing

Answer 19

Thanks I really appreciate it! I'm brand new to this and I think I should change my name, do you know how to do that too?

Answer 20

have no idea You might have to send them something or what not

Answer 21

Thanks, I need to get back to studying now, again I thank you for the help