The implicit function:

`x\ cos\ 2y+sin\ x\ cos\ y=1`

We differentiate each term from left to right:

`x(-2\ sin\ 2y)((dy)/(dx))` `+(cos\ 2y)(1)` `+sin\ x(-sin\ y(dy)/(dx))` `+cos\ y\ cos\ x`

`=0`

So

`(-2x\ sin\ 2y-sin\ x\ sin\ y)((dy)/(dx))` `=-cos\ 2y-cos\ y\ cos\ x`

Solving for `dy/dx` gives us:

`(dy)/(dx)` `=(-cos\ 2y-cos\ y\ cos\ x)/(-2x\ sin\ 2y-sin\ x\ sin\ y)`

`= (cos\ 2y+cos\ x\ cos\ y)/(2x\ sin\ 2y+sin\ x\ sin\ y)`

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