Question

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

Answer 1

Let's use the chain rule, by naming `u=(sinx)/(1+cosx)`.

The chain rule states that

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

Thus,

`(dy)/(du)=2u`

`(du)/(dx)` - here, we have to apply quocient rule, which states that

Be `y=(f(x))/(g(x))`, `(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(f(x))^2`

`(du)/(dx)=(((cosx)(1+cosx))-(sinx)(-sinx))/(1+cosx)^2`

`(du)/(dx)=((cos^2x+cosx)+sin^2x)/(1+cosx)^2`

`(du)/(dx)=(color(green)(cos^2x+sin^2x)+cosx)/(1+cosx)^2`

`(du)/(dx)=cancel(1+cosx)/(1+cosx)^cancel2`

`(du)/(dx)=1/(1+cosx)`

Now, combining `(du)/(dx)` and `(dy)/(du)`:

`(dy)/(dx)=(2u)(1/(1+cosx))`

`(dy)/(dx)=2(sinx/(1+cosx))(1/(1+cosx))`

`(dy)/(dx)=color(green)((2sinx)/(1+cosx)^2)`