Question

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

Answer 1

`f'(x) = -1/(1+sinx)`

Explanation:

Use the quotient rule to get the derivative. The quotient rule says that if `f(x) = g(x)/(h(x))` the derivative `f'(x) = (g'(x)h(x) - g(x)h'(x))/(h^2(x))`
In this case `g(x) = cosx` and `h(x) = (1+sinx)`
`g'(x) =-sinx`
`h'(x) = cosx`
`f'(x) = (-sinx(1+sinx) - cosx*cosx)/(1+sinx)^2`
`f'(x) = (-sinx-sin^2x -cos^2x)/(1+sinx)^2`
`f'(x) =-(sinx+sin^2x +cos^2x)/(1+sinx)^2`
We know that `sin^2x +cos^2x = 1` so the expression becomes
`f'(x) = -(sinx+1)/(1 + sinx)^2`
`f'(x) = -1/(1+sinx)`