This is an implicit function.

3 cot(x + y) = cos y2

For the left hand side, we put u = x + y.

Differentiating 3 cot u gives us:

`3(-csc^2 u)((du)/(dx))`

Substituting for `u` and performing the `(du)/(dx)` part gives us:

`-3 csc^2(x+y)(1+(dy)/(dx))`

On the right hand side, we let u = y2. Differentiating `cos u` gives us:

`(-sin u)((du)/(dx))`

Substituting for `u` and performing the `(du)/(dx)` part gives us:

`(-sin y^2)(2y(dy)/(dx))`

We put both sides together:

`-3 csc^2(x+y)(1+(dy)/(dx))` `=(-sin y^2)(2y(dy)/(dx))`

Expanding gives:

`-3 csc^2(x+y)` `-3 csc^2(x+y)(dy)/(dx)` `=-2y sin y^2(dy)/(dx)`

`2y sin y^2(dy)/(dx)`

to both sides:

`-3 csc^2(x+y)` `-3 csc^2(x+y)(dy)/(dx)` `+2y sin y^2(dy)/(dx)` `=0`

`3 csc^2(x+y)`

to both sides:

`-3 csc^2(x+y)(dy)/(dx)` `+2y sin y^2(dy)/(dx)` `=3 csc^2(x+y)`

Factoring out the dy/dx term:

`[2y sin y^2-3 csc^2(x+y)](dy)/(dx)` `=3 csc^2(x+y)`

This gives us:

`(dy)/(dx)`

`=(3 csc^2(x+y))/(2y sin y^2` `-3 csc^2(x+y)`

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