Question

Usando a regra da cadeia, determinar a derivada primeira das funções abaixo
f (x) = 2x /(3x −1)^1/ 2
Resposta : f '( x) = (9x − 2) /(3x −1)^1/ 2

não consegui chegar nesta resposta

Answer 1

$\\f(x)=\frac{2x}{(3x-1)^{\frac{1}{2}}}\\\\\\y=\frac{2x}{(\sqrt{3x-1})^2}\\\\\\y^2=\frac{4x^2}{3x-1}\\\\3xy^2-y^2=4x^2\\\\3dx\cdot\,y^2+3x\cdot2ydy-2ydy=8xdx$

$\\3dx\cdot\,y^2+3x\cdot2ydy-2ydy=8xdx\\\\6xy\cdot\,dy-2y\cdot\,dy=8x\cdot\,dx-3y^2\cdot\,dx\\\\2y(3x-1)dy=(8x-3y^2)dx\\\\2\cdot\frac{2x}{\sqrt{3x-1}}\cdot(3x-1)dy=\left(8x-3\cdot\frac{4x^2}{3x-1}\right)dx\\\\\\\frac{4x}{\sqrt{3x-1}}\cdot(3x-1)dy=4x\left(2-\frac{3x}{3x-1}\right)dx$

$\\\frac{1}{\sqrt{3x-1}}\cdot(3x-1)dy=1\left(\frac{6x-2-3x}{3x-1}\right)dx\\\\\\\frac{3x-1}{\sqrt{3x-1}}dy=\frac{3x-2}{3x-1}dx\\\\\\\frac{dy}{dx}=\frac{3x-2}{3x-1}\cdot\frac{\sqrt{3x-1}}{3x-1}\\\\\\\frac{dy}{dx}=\frac{(3x-2)(3x-1)^{\frac{1}{2}}}{(3x-1)^2}\\\\\\\frac{dy}{dx}=\frac{(3x-2)}{(3x-1)^2\cdot(3x-1)^{-\frac{1}{2}}}\\\\\\\boxed{\frac{dy}{dx}=\frac{(3x-2)}{(3x-1)^{\frac{3}{2}}}}$