Question

Use implicit differentiation and solve
y=cos(x-y)

Answer 1

\[[\cos(x-y)]'=(x-y)'(-\sin(x-y))\]

Answer 2

how would you isolate for y'

Answer 3

\[\frac{dy}{dx}=\left(\frac{dy}{dx}-1\right)\sin\left(x-y\right)\]\[\frac{dy}{dx}=\frac{dy}{dx}\sin\left(x-y\right)-\sin\left(x-y\right)\]\[1=\sin\left(x-y\right)-\frac{dx}{dy}\sin\left(x-y\right)\]\[1+\frac{dx}{dy}\sin\left(x-y\right)=\sin\left(x-y\right)\]\[\frac{dx}{dy}\sin\left(x-y\right)=\sin\left(x-y\right)-1\]\[\frac{dx}{dy}=1-\csc\left(x-y\right)\]\[\frac{dy}{dx}=\frac{1}{1-\csc\left(x-y\right)}\]

Answer 4

Why is everyone so deathly quiet? Am I wrong?

Answer 5

Im sorry. Im a bit of a slow learner.... I dont quite follow

Answer 6

Sure, I'll do it step by step. Give me a while to type stuff up, though.

Answer 7

Or if Zarkon is going to correct me, I will stand my ground. :P

Answer 8

your answer is correct, but...

I would avoid using \(\displaystyle\frac{dx}{dy}\), not that it is wrong, but it might be confusing to someone who is not really comfortable with derivatives.

Answer 9

I would keep the answer in the form \[\frac{\sin(x-y)}{\sin(x-y)-1}\]

Answer 10

I honestly will say right now that I don't know how to do this without using the property that \(dx,dy\) are seperate mathemtical entities that can be inverted.

It might be more appropriate that you pick up from here, because I don't know how else to explain it.

Answer 11

Thanks @Zarkon :P

Answer 12

Oh, right, I can subtract from both sides two parts that contain the derivative, and factor it out. Okay, ignore me. XD

Answer 13

Hey, I just had a test. My brain isn't working at full speed.

Answer 14

from here...
\[\frac{dy}{dx}=\frac{dy}{dx}\sin\left(x-y\right)-\sin\left(x-y\right)\]

\[\frac{dy}{dx}-\frac{dy}{dx}\sin\left(x-y\right)=-\sin\left(x-y\right)\]

\[\frac{dy}{dx}(1-\sin\left(x-y\right))=-\sin\left(x-y\right)\]

\[\frac{dy}{dx}=\frac{-\sin\left(x-y\right)}{1-\sin\left(x-y\right)}\]

\[\frac{dy}{dx}=\frac{\sin\left(x-y\right)}{\sin\left(x-y\right)-1}\]

Answer 15

:)

Answer 16

@bobobobobb You should give Zarkon the medal. ;)

Answer 17

no

Answer 18

Okay, but one question: how did you get dy/dx sin(x-y) - sin(x-y)

Answer 19

badreferences distributed the sin(x-y) over the two terms

Answer 20

look at badreferences's work then mine. I just continued his work after his second step

Answer 21

y=cos(x-y) ,, zarcons answer is correct.. use or let u=(x-y)
so that y=cos u,
then dy/dx= d(cos u)/dx
= - sin u du/dx,,, but du/dx=(1-dy/dx)
=-[sin u][1-dy/dx]
=-sin u +(dy/dx)sin u,,collecting terms gives...
dy/dx-(dy/dx)sin u=-sin u
dy/dx(1-sin u) = -sin u
dy/dx= -sin u/(1-sin u) or .....................du/dx= -sin (x-y)/(1-sin (x-y)) .ans
dy/dx= sin u/(sin u +1) but u=(x-y) subst..
dy/dx=sin(x-y)/[sin(x-y) +1] ans

Answer 22

im sorry that is
dy/dx=sin(x-y)/[sin(x-y) -1] ans

Answer 23

if you want you can do it the really slick way and compute

\[\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}\]


where f(x,y)=y-\cos(x-y)

;)

Answer 24

\[f(x,y)=y-\cos(x-y)\]

Answer 25

I finally understand. Thank you so much guys. I entered a calculus course where I am too young and I dont understand a thing. But thanks for your time guys.