Question

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

Answer 1

`(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`