Question

Question #232e8

Answer 1

`f'(x)= -cos x`

Explanation:

Given `f(x) = (cosx)^2/(1+sinx)`
*Note `(cosx)^2= cos^2x`

You can find the derivative using the quotient rule
`f(x) = (h(x))/g(x)` then
`"f'(x) = (g(x)*h'(x) -h(x)*g'(x))/[g(x)]^2` `" " " " (1)`

1. Let `h(x) = (cosx)^2` then

Note: to differentiate `h(x)` we need to use the power and chain rule

`h'(x) = 2 (cosx)^(2-1)* 1(-sin x)`
`color(red)(h'(x) = -2sinx cos x)`

1. Let `g(x) = 1+ sin x`
   `color(blue)(g'(x) = cos x *1) =>color(blue)( cosx)`

2. Let's substitute into (1) to get
   `f'(x) =((1+ sinx)color(red)((-2sinx cos x)) -cos^2x* color(blue)(cosx))/(1+sinx)^2`

Factot out the greatest common factor `"-cos x`

`=(-cosx[(1+sinx)(2sinx)+cos^2x])/(1+sinx)^2`

Rewrite `cos^2 x= 1-sin^2x` using Trigonometric Pythagorean identity

`=(-cosx(2sinx +2sin^2x+color(green)(1-sin^2x)))/(1+sinx)^2`

`=(-cosx(sin^2x +2sinx +1))/(1+sinx)^2`

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

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

`=(-cosxcancel((sinx+1)^2))/cancel((1+sinx)^2)`

`f'(x)= -cos x`

I hope this help.