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.