Implicit Differentiation of Trig Functions

[suicoted]

Find dy/dx by imp. differentiation. I can't seem to get the right answer for
(a) y = cox (x-y)
The answer in the book is: -sin(x-y)/1-sin(x - y)

My method:

y' = [cos x cos y + sin x sin y]' {I used trig identity}
= (- sin x) ( - sin y ) + ( cos x ) ( cos y ) {I took derivative of each}


(b) csc (x - y) + sec (x + y) = x (I have no clue what to do).
I'm actually more interested in the process of doing it than the solution. Help would be appreciated .

 

[pka]

(a) y = cos(x-y)
The answer in the book is: -sin(x-y)/1-sin(x - y)
My method: {I used trig identity} as waste of time!

y’=-sin(x-y)(1-y’)=-sin(x-y)+y’sin(x-y).
Now you solve for y’ !

 

[soroban]

Hello, suicoted!

(b) csc (x - y) + sec (x + y) = x
(I have no clue what to do). . . . . . really?

How about differentiating csc(x - y) . . . by the Chain Rule?
. . . - csc(x-y) cot(x-y) (1 - y')

Then differentiating sec(x + y) . . . by the Chain Rule?
. . . sec(x + y) tan(x + y) (1 + y')

The differentiate the x on the right side: 1


We have: . -csc(x-y) cot(x-y) (1 - y') + sec(x+y) tan(x + y) (1 - y') .= .1

Then solve for y' . . .