Question #232e8
1 Answer
Explanation:
Given
*Note
You can find the derivative using the quotient rule
- Let
#h(x) = (cosx)^2# then
Note: to differentiate
#h'(x) = 2 (cosx)^(2-1)* 1(-sin x)#
#color(red)(h'(x) = -2sinx cos x)#
-
Let
#g(x) = 1+ sin x#
#color(blue)(g'(x) = cos x *1) =>color(blue)( cosx) # -
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
#=(-cosx[(1+sinx)(2sinx)+cos^2x])/(1+sinx)^2#
Rewrite
#=(-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)#
I hope this help.