how can I calculate this deriivative:

Question

1/(sin(x-sinx))?

Answer 1

Chain rule:

d/dx [1/(sin(x-sinx))] =

-1/[sin^2(x-sin(x)] *
d/dx [sin(x-sin(x))] =

-cos(x-sin(x))/[sin^2(x-sin(x)]*
d/dx [x-sin(x)] =

[cos(x)-1]cos(x-sin(x))/[sin^2(x-sin(x)]