Question

What is the derivative of `sinx/(1+cosx)`?

Answer 1

Use the division's derivative formula:

For a given function g: `g=u/v` for `u` and `v!=0` other functions, the derivative of g is found as;

`g'=(u'v-uv')/v^2`

If we apply it to our case:

`f'(x)=((sinx)'(1+cosx)-sinx(1+cosx)')/(1+cosx)^2= (cosx(1+cosx)+sinxsinx)/(1+cosx)^2=(cosx+cos^2x+sin^2x)/(1+cosx)^2`

Using now the trigonometric rule: `cos^2x+sin^2x=1` we finally get:

`f'(x)=(1+cosx)/(1+cosx)^2=1/(1+cosx)`