anonymous
  • anonymous
Show the steps of the derivative of cot(x+y) +cotx+coty = 0 Thanks:)
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
Implicit diff. or partial derivative?
anonymous
  • anonymous
I dont know how:(
anonymous
  • anonymous
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.

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anonymous
  • anonymous
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.
anonymous
  • anonymous
ohh i think it is implicit
anonymous
  • anonymous
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
anonymous
  • anonymous
Heres what I did, but I get it wrong:
anonymous
  • anonymous
cot(x+y) + cotx+ coty = 0 \[-\csc^2(x+y)(1) + -\csc^2(x) + dy/dx -\csc^2(y) = 0 \] \[(-\csc^2(x)/ (\csc^2(x+y) + \csc^2(y)\]
anonymous
  • anonymous
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)).
anonymous
  • anonymous
If it were 1+dy/dx, what difference is it? What does dy/dx by itself mean?
anonymous
  • anonymous
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.
anonymous
  • anonymous
ohh i think i get it now. Does that mean dy/dx always goes with y then?
anonymous
  • anonymous
Yes!
anonymous
  • anonymous
Alright thanks for the time and patience brainshot3:D
anonymous
  • anonymous
You will be the first person I fan
anonymous
  • anonymous
=D

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