Derivative of trigonometric function;

y = 2sinx/(4-3sinx)

Find dy/dx.

I keep coming to;
[(2cosx)(4-3sinx)-(2sinx)(-3cosx)]/(4-…

But evidently the answer is 8cosx/(4-3sinx)^2.

Could someone show me how one gets to that answer?

I think you got the right answer, you just didn't simplify.

Note that (2cosx)(4-3sinx)-(2sinx)(-3cosx) expands to give 8cosx

EDIT

(2cosx)(4-3sinx) = 2cos(x)*4 + 2cos(x)*-3sin(x) = 8 cos(x) - 6 sin(x) cos(x)
(2sinx)(-3cosx) = -6 cos(x) sin(x)

therefore (2cosx)(4-3sinx)-(2sinx)(-3cosx) = 8 cosx - 6 cosx sinx + 6 cosx sinx
= 8 cosx