Question

What is the derivative of `( cos (x) ) / ( 2 + sin (x) )`?

Answer 1

`f'(x)=\frac{-2\sin x-1}{(2+\sin x)^2}`

Explanation:

Given function:

`f(x)=\frac{\cos x}{2+\sin x}`

Differentiating above function w.r.t. `x` using quotient rule as follows

`d/dxf(x)=d/dx(\frac{\cos x}{2+\sin x})`

`f'(x)=\frac{(2+\sin x)d/dx(\cos x)-\cos xd/dx(2+\sin x)}{(2+\sin x)^2}`

`=\frac{(2+\sin x)(-\sin x)-\cos x(\cos x)}{(2+\sin x)^2}`

`=\frac{-2\sin x-\sin^2x-\cos ^2x}{(2+\sin x)^2}`

`=\frac{-2\sin x-(\sin^2x+\cos ^2x)}{(2+\sin x)^2}`

`=\frac{-2\sin x-1}{(2+\sin x)^2}`

Answer 2

`-(2sinx+1)/(2+sinx)^2`

Explanation:

We have a quotient here, so we can find the derivative using quotient rule which, if you forgot, is:

`d/dx(g(x)/(h(x))) = (g'(x)*h(x) - h'(x)*g(x))/(h^2(x))`

So let's apply it:

`f(x)=cosx/(2+sinx)`

`f'(x)=\frac{(2+\sin x)d/dx(\cos x)-(\cos x)d/dx(2+\sin x)}{(2+\sin x)^2}`

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

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

`f'(x)=(-2sinx-(color(red)(sin^2x+cos^2x)))/(2+sinx)^2`

To simplify this further, we can use the pythagorean identity:
`color(red)(sin^2x+cos^2x=1`

So:

`f'(x)=(-2sinx-1)/(2+sinx)^2` or `-(2sinx+1)/(2+sinx)^2`