1. ## implicit different.

coty = x - y

i worked it to...
dy/dx = 1/(1-csc²y)

2. Originally Posted by nuclear123
coty = x - y

i worked it to...
-csc²y*(dy/dx) = 1 - (dy/dx)

unsure of how to finish this haveing dy/dx on both sides of the equation? if i move them both over to one side i get -csc²y*(dy/dx) + (dy/dx) = 1 but still unsure on how to finish it? any help -thx
From your last step you can factor out a dy/dx

$-\csc ^2(y)\, \frac{dy}{dx} + \frac{dy}{dx} = \frac{dy}{dx}\,(1- \csc ^2(y))$

Edit, if it helps: $1- \csc^2(y) = - \cot^2(y)$

3. ## s

i got that but over 1....not sure what i did wrong i added the dy/dx over and it left 1 on the right then divided

4. ## bump

bump*
anyone explain why i cant work it the way i did? i dont get how u can factor out a dy/dx..

5. Originally Posted by nuclear123
coty = x - y

i worked it to...
dy/dx = 1/(1-csc²y)

$\frac{1}{1-\csc^2y}=\frac{1}{-\cot^2y}=-\tan^2y$

Originally Posted by nuclear123
bump*
anyone explain why i cant work it the way i did? i dont get how u can factor out a dy/dx..
ab + b = b(a + 1)

$-\csc ^2y\, \frac{dy}{dx} + \frac{dy}{dx} =1$

$\frac{dy}{dx}\,(- \csc ^2y+1)=1$

$\frac{dy}{dx}=\frac{1}{- \csc ^2y+1}$