Question

calcule a derivada da função y=ln(1+x/1-x)

Answer 1

Explicação passo-a-passo:

Temos que:

$\sf y=ln~u~\Rightarrow~y'=\dfrac{u'}{u}$

Assim:

$\sf y=ln~\left(\dfrac{1+x}{1-x}\right)$

$\sf y'=\left(\dfrac{1+x}{1-x}\right)'\cdot\dfrac{1-x}{1+x}$

$\sf y'=\dfrac{(1+x)'\cdot(1-x)-(1-x)'\cdot(1+x)}{(1-x)^2}\cdot\dfrac{1-x}{1+x}$

$\sf y'=\dfrac{1\cdot(1-x)-(-1)\cdot(1+x)}{(1-x)^2}\cdot\dfrac{1-x}{1+x}$

$\sf y'=\dfrac{1-x+1-x}{(1-x)^2}\cdot\dfrac{1-x}{1+x}$

$\sf y'=\dfrac{2}{(1-x)^2}\cdot\dfrac{1-x}{1+x}$

$\sf y'=\dfrac{2}{(1-x)\cdot(1+x)}$