Question

What is the derivative of `y=(3(1-sinx))/(2cosx)`?

Answer 1

Use the rule `(u/v)' = (u'v-uv')/v^2`.

Here, `u(x) = 3(1- sin(x))` and `v(x) = 2 cos(x)`.

So, the derivative is

`(-3cos(x) times (2 cos(x)) - 3(1-sin(x))times (-2 sin(x)))/(4 cos^2(x))`

or easier `(-6cos^2(x) + 6(sin(x) - sin^2(x)))/(4cos^2(x))`

But remember `cos^2(x)+sin^2(x)=1`, you can write

`(-6 + 6sin(x))/(4cos^2(x)) = 3/2 (-1+sin(x))/cos^2(x)`