Find dy/dx by implicit differentiation.
y=cos(x-y)

Question

Find dy/dx by implicit differentiation.
y=cos(x-y)

anonymous · Answer

I got this for now.
y' -sin(x-y)(1-y')

zepdrix · Answer

Mmmm ok looks good so far!
Having trouble solving for y'?

anonymous · Answer

ok

anonymous · Answer

Why do we add sin(x-y) on each side for?

anonymous · Answer

Yeah, I didn't quite get it...

anonymous · Answer

How did you get ...=y' +sin(x-y)y' ?

zepdrix · Answer

Maybe this sine is confusing? Let's call it something else for now.$$\Large\rm y'=-\color{orangered}{\sin(x-y)}(1-y')$$$$\Large\rm y'=-\color{orangered}{A}(1-y')$$I'm distributing A to each term in the brackets,$$\Large\rm y'=\color{orangered}{A}y'-\color{orangered}{A}$$

zepdrix · Answer

Then we'll subtract Ay' from each side,$$\Large\rm y'-Ay'=\cancel{Ay'}-A\cancel{-Ay'}$$

zepdrix · Answer

$$\Large\rm y'-Ay'=-A$$$$\Large\rm y'-\sin(x-y)y'=-\sin(x-y)$$

zepdrix · Answer

Hmm what do you think?
Did the A's make it MORE confusing? 0_o lol

anonymous · Answer

Oh I see! I got it :)

zepdrix · Answer

Oh I made a mistake the first time didn't I? >.<
I had a minus on the left side, it was -y' over there.
Led to a mistake.

This final one though is ok.

zepdrix · Answer

Lemme erase those old notes a sec

zepdrix · Answer

Understand how to isolate the y' from there? :o

anonymous · Answer

Yes, I got it now! Thank you! :)

zepdrix · Answer

cool c: