Dezembro 23, 2019

por Viviyan

Qual é a derivada do $arctan (2 x)$ $arctan (2x)$ ?

Responda:

$\frac{2}{1 + 4 x^{2}}$ $2/(1+4x^2)$

Explicação:

using $\frac{d}{d x} ({tan}^{- 1} x) = \frac{1}{1 + x^{2}}$ $d/dx (tan^-1x) = 1/(1+x^2)$

differentiating using the $chain rule$ $color(blue)(" chain rule ")$

here x = 2x , hence

$\Rightarrow \frac{d}{d x} ({tan}^{- 1} 2 x) = \frac{1}{1 + {(2 x)}^{2}} \frac{d}{d x} (2 x)$ $rArr d/dx(tan^-1 2x) = 1/(1+(2x)^2) d/dx(2x)$

$= \frac{1}{1 + 4 x^{2}} .2 = \frac{2}{1 + 4 x^{2}}$ $= 1/(1+4x^2) .2 = 2/(1+4x^2)$