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.