Question

Show the steps of the derivative of cot(x+y) +cotx+coty = 0
Thanks:)

Answer 1

Implicit diff. or partial derivative?

Answer 2

I dont know how:(

Answer 3

Are you in Calc 3 or 1? I'm just not sure if you are doing the partial derivative of x or y or implicit diff.

Answer 4

xP not sure what that is. Are you doing implicit differentiation?? It has the dy/dx, partial derivatives are when you treat y as a constant.

Answer 5

ohh i think it is implicit

Answer 6

Ok, so you would take the derivative of x+y because they are inside Cot first and you get 1+ (dy/dx), then you multiply by the derivative of Cot with x+y within it. Now, you just need to take the derivative of the other cotx and Cot(y). Cot(y) is different because you follow the chain rule and the derivative of y is (dy/dx), not 1, so you would have (dy/dx)* derivative of Cot, then solve for dy/dx

Answer 7

Heres what I did, but I get it wrong:

Answer 8

cot(x+y) + cotx+ coty = 0
\[-\csc^2(x+y)(1) + -\csc^2(x) + dy/dx -\csc^2(y) = 0 \]
\[(-\csc^2(x)/ (\csc^2(x+y) + \csc^2(y)\]

Answer 9

For the first part you only took the derivative of x, not x+y. You should have in parentheses, instead of just (1), (1+(dy/dx)).

Answer 10

If it were 1+dy/dx, what difference is it? What does dy/dx by itself mean?

Answer 11

dy/dx is the derivative of y in implicit differentiation, so you would need to distribute the (1+(dy/dx)), then put all terms with dy/dx on one side, factor out the dy/dx and put dy/dx on one side and everything else on the other.

Answer 12

ohh i think i get it now. Does that mean dy/dx always goes with y then?

Answer 13

Yes!

Answer 14

Alright thanks for the time and patience brainshot3:D

Answer 15

You will be the first person I fan

Answer 16

=D