Question

Socorro genteeeeeeeeeeeeeeeee

Answer 1

ta tchau volto amanha ta

Answer 2

$\dfrac{df(x)}{dx} = \dfrac{-\dfrac{d}{dx}\left[sen\left(\dfrac{x+1}{x-1}\right)\right]}{sen^2\left(\dfrac{x+1}{x-1}\right)}$

Enquanto:

$\dfrac{d}{dx}\left[sen\left(\dfrac{x+1}{x-1}\right)\right] = \cos\left(\dfrac{x+1}{x-1}\right) \cdot \dfrac{1 \cdot (x-1) - (x+1) \cdot 1}{(x-1)^2}$

$\dfrac{d}{dx}\left[sen\left(\dfrac{x+1}{x-1}\right)\right] = \cos\left(\dfrac{x+1}{x-1}\right) \cdot \dfrac{-2}{(x-1)^2}$

Substituindo:

$\dfrac{df(x)}{dx} = \dfrac{-\cos\left(\dfrac{x+1}{x-1}\right) \cdot \dfrac{-2}{(x-1)^2}}{sen^2\left(\dfrac{x+1}{x-1}\right)}$

Sabendo que:

$\dfrac{cos(x)}{sen(x)} = cotg(x)$,

teremos:

$\dfrac{df(x)}{dx} = \dfrac{2}{(x-1)^2} \cdot cotg\left(\dfrac{x+1}{x-1} \right) \cdot\dfrac{1}{sen\left(\dfrac{x+1}{x-1}\right)}$

Alternativa A