Question #232e8

1 Answer
Dec 12, 2015

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.