How do you find the derivative of # F(x) = (sinx/(1+cosx))^2#?

1 Answer
Mia
Aug 2, 2016

#(2sinx)/(1+cosx)^2#

Explanation:

The derivative of a power is stated as follows:
#( (u)^n)'=n*u*u'#

Applying this property the derivative of the given power is :

#color(blue)((sinx/(1+cosx)^2)'=2(sinx/(1+cosx))(sinx/(1+cosx))')#

Let's find# (sinx/(1+cosx))'#:
The derivative of the quotient
# (u/v)'=(u'v-v'u)/v^2#

To apply the derivative of a quotient on (sinx)/(1+cosx) we've to find :

#(sinx)'=cosx#
#(1+cosx)'= -sinx#
So,
#color(blue)((sinx/(1+cosx))')#
#=((sinx)'(1+cosx)-(1+cosx)'sinx)/(1+cosx)^2#
#=(cosx(1+cosx)-(-sinx)(sinx))/(1+cosx)^2#

#=(cosx+(cosx)^2+(sinx)^2)/(1+cosx)^2#

Knowing that# (cosx)^2+(sinx)^2=1#
#=(cosx+1)/(1+cosx)^2#

Simplifying by #1+cosx#
Therefore,
#(sinx /(1+cosx))'=1/(1+cosx)#

So ,the derivative of the given power:
#((sinx/(1+cosx))^2)'#
#=2(sinx/(1+cosx))(sinx/(1+cosx))'#
#=2(sinx/(1+cosx))(1/(1+cosx))#
#=(2sinx)/(1+cosx)^2#

Related questions
See all questions in Summary of Differentiation Rules
Impact of this question
5520 views around the world
You can reuse this answer
Creative Commons License