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

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Answer 1 · 2018-04-04T15:44:40

$- 2 \frac{sin (2 t)}{{cos (2 t)}^{2} + 1}$ $-2sin(2t)/(cos(2t)^2+1)$

The derivative of ${tan}^{- 1} (x)$ $tan^-1(x)$ is $\frac{1}{x^{2} + 1}$ $1/(x^2+1)$

when we substitute $cos (2 t)$ $cos(2t)$ for x we get

$\frac{1}{{cos (2 t)}^{2} + 1}$ $1/(cos(2t)^2+1)$

Then we apply the chain rule for cos(2t)

$\frac{1}{{cos (2 t)}^{2} + 1} \cdot - 2 sin (2 t)$ $1/(cos(2t)^2+1) * -2sin(2t)$

Our final answer is

$- 2 \frac{sin (2 t)}{{cos (2 t)}^{2} + 1}$ $-2sin(2t)/(cos(2t)^2+1)$

What is the derivative of arctan(cos2t)arctan (cos 2t)?