Question

What is the derivative of `arctan (cos 2t)`?

Answer 1

`-(2sin2t)/(1+cos^2 2t`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"Given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`rArrd/dt(arctan(cos2t))`

`=1/(1+(cos2t)^2)xxd/dt(cos2t)xxd/dt(2t)`

`=1/(1+cos^2 2t)xx-sin2txx2`

`=-(2sin2t)/(1+cos^2 2t)`

Answer 2

`(dy)/(dt)=-(2sin2t)/(1+cos^2 2t)`

Explanation:

We know that,

`color(red)((1)d/(dx)(tan^-1x)=1/(1+x^2)`

`color(blue)((2)d/(dx)(cosx)=-sinx`

Let,

`y=f(t)=tan^-1(cos2t)`

Diff, w .r .to. `color(violet)"(t)"`

`(dy)/(dt)=color(red)(1/(1+(cos2t)^2))xxd/(dt)(cos2t)..color(red)(toApply(1)`

`=>(dy)/(dt)=1/(1+cos^2 2t)xx(color(blue)(-sin2t))d/(dt)(2t) color(blue)(to Apply(2)`

`=>(dy)/(dt)=-(sin2t)/(1+cos^2 2t)xx2`

`i.e. (dy)/(dt)=-(2sin2t)/(1+cos^2 2t)`