Question

Seja y = (Raiz quadrada de x^2 - 1 )/ x, determine dy/dx. (Assunto: Derivadas)

Answer 1

Devemos saber que:

$h(x)=\dfrac{f (x)}{g(x)}\Longrightarrow h'(x)=\dfrac{f'(x)\cdot g(x)-g'(x)\cdot f(x)}{[g(x)]^2}$

Usando na expressão dada:

$y=\dfrac{\sqrt{x^2-1}}{x}=\dfrac{(x^2-1)^{\frac{1}{2}}}{x}\\\\ \dfrac{dy}{dx}=\dfrac{[(x^2-1)^{\frac{1}{2}}]'\cdot x-[x]'\cdot(x^2-1)^{\frac{1}{2}}}{x^2}\\\\ \dfrac{dy}{dx}=\dfrac{\frac{1}{2}[(x^2-1)^{-\frac{1}{2}}]\cdot 2x\cdot x-1\cdot(x^2-1)^{\frac{1}{2}}}{x^2}\\\\ \dfrac{dy}{dx}=\dfrac{x^2(x^2-1)^{-\frac{1}{2}}-(x^2-1)^{\frac{1}{2}}}{x^2}$

Multiplicando a expressão da direita por $(x^2-1)^{\frac{1}{2}}$ no numerador e no denominador:

$\dfrac{dy}{dx}=\dfrac{x^2-(x^2-1)}{x^2(x^2-1)^{\frac{1}{2}}}\\\\ \dfrac{dy}{dx}=\dfrac{1}{x^2(x^2-1)^{\frac{1}{2}}}\\\\ \boxed{\dfrac{dy}{dx}=\dfrac{1}{x^2\sqrt{x^2-1}}}$