Question

Resolva a seguinte derivada
y= $\sqrt{\frac{1+x}{1-x} }$

Answer 1

Calcular a derivada com relação a x da função

    $\mathsf{y=\sqrt{\dfrac{1+x}{1-x}}}$

Reescrevendo em forma de potência:

    $\mathsf{\Longleftrightarrow\quad y=\left(\dfrac{1+x}{1-x}\right)^{\!1/2}}$

Derive usando a regra da potência:

    $\mathsf{\Longrightarrow\quad \dfrac{dy}{dx}=\dfrac{d}{dx}\!\left[\left(\dfrac{1+x}{1-x}\right)^{\!1/2}\right]}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{1}{2}\left(\dfrac{1+x}{1-x}\right)^{\!(1/2)-1}\cdot \dfrac{d}{dx}\!\left(\dfrac{1+x}{1-x}\right)}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{1}{2}\left(\dfrac{1+x}{1-x}\right)^{\!-1/2}\cdot \dfrac{d}{dx}\!\left(\dfrac{1+x}{1-x}\right)}$

Agora aplique a regra para a derivada do quociente:

    $\mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{1}{2}\left(\dfrac{1+x}{1-x}\right)^{\!-1/2}\cdot \dfrac{\frac{d}{dx}(1+x)\cdot (1-x)-(1+x)\cdot \frac{d}{dx}(1-x)}{(1-x)^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{1}{2}\left(\dfrac{1+x}{1-x}\right)^{\!-1/2}\cdot \dfrac{1\cdot (1-x)-(1+x)\cdot (-1)}{(1-x)^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{1}{2}\left(\dfrac{1+x}{1-x}\right)^{\!-1/2}\cdot \dfrac{1-\diagup\!\!\!\! x+(1+\diagup\!\!\!\! x)}{(1-x)^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{1}{\diagup\!\!\!\! 2}\left(\dfrac{1+x}{1-x}\right)^{\!-1/2}\cdot\, \dfrac{\diagup\!\!\!\! 2}{(1-x)^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\left(\dfrac{1+x}{1-x}\right)^{\!-1/2}\cdot \dfrac{1}{(1-x)^2}}$

    $\mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\left(\dfrac{1-x}{1+x}\right)^{\!1/2}\cdot \dfrac{1}{(1-x)^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\sqrt{\dfrac{1-x}{1+x}}\cdot \dfrac{1}{(1-x)^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{1}{(1-x)^2}\,\sqrt{\dfrac{1-x}{1+x}}\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)

Answer 2

Explicação passo-a-passo:

.

Cálculo da derivada :

$\mathsf{y~=~\sqrt{\dfrac{1+x}{1-x}} }$

$\mathsf{y~=~\Bigg(\dfrac{1+x}{1-x} \Bigg)^{\frac{1}{2}} }$

$\mathsf{y~=~\Big(f(x)\Big)^n }$

$\mathsf{y'~=~n.\Big(f(x) \Big)^{n-1}.f'(x) }$

$\mathsf{y'~=~\dfrac{1}{2}.\Bigg(\dfrac{1+x}{1-x} \Bigg)^{\frac{1}{2}-1}.\Big(\dfrac{1+x}{1-x}\Big)' }$

$\mathsf{y'~=~\dfrac{1}{2}.\Bigg(\dfrac{1+x}{1-x} \Bigg)^{-\frac{1}{2}} . \dfrac{(1+x)'.(1-x)-(1+x).(1-x)'}{(1-x)^2} }$

$\mathsf{y'~=~\dfrac{1}{2}.\dfrac{1}{\Big(\frac{1+x}{1-x}\Big)^{\frac{1}{2}}}.\dfrac{(1-x)-(-1)(1+x)}{(1-x)^2} }$

$\mathsf{y'~=~\dfrac{1}{2}.\dfrac{1}{\sqrt{\frac{1+x}{1-x}}}.\dfrac{1-x +x+1}{(1-x)^2} }$

$\boxed{\boxed{\mathsf{y'~=~\dfrac{1}{2\sqrt{\frac{1+x}{1-x}}}.\Bigg(\dfrac{2}{(1-x)^2} \Bigg) }}}}$

Espero ter ajudado bastante!):