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

Apr 21, 2018

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

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{Given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\Rightarrow \frac{d}{\mathrm{dt}} \left(\arctan \left(\cos 2 t\right)\right)$

$= \frac{1}{1 + {\left(\cos 2 t\right)}^{2}} \times \frac{d}{\mathrm{dt}} \left(\cos 2 t\right) \times \frac{d}{\mathrm{dt}} \left(2 t\right)$

$= \frac{1}{1 + {\cos}^{2} 2 t} \times - \sin 2 t \times 2$

$= - \frac{2 \sin 2 t}{1 + {\cos}^{2} 2 t}$

Apr 21, 2018

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{2 \sin 2 t}{1 + {\cos}^{2} 2 t}$

#### Explanation:

We know that,

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

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

Let,

$y = f \left(t\right) = {\tan}^{-} 1 \left(\cos 2 t\right)$

Diff, w .r .to. $\textcolor{v i o \le t}{\text{(t)}}$

(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)

$\implies \frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{\sin 2 t}{1 + {\cos}^{2} 2 t} \times 2$

$i . e . \frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{2 \sin 2 t}{1 + {\cos}^{2} 2 t}$