calcule a derivada de f(x)=(x+1)/(x-1)

Você pode resolver pela Regra do Quociente: f'(x) = [(x+1)' * (x-1) - (x+1) * (x-1)'] / (x-1)²

R.: -2 / (x-1)²

Abs

Seja:

\(\frac{d}{dx}\left(\frac{\left(x+1\right)}{x-1}\right)\)

Vamos aplicar a regra do quociente: \((\frac{f}{g})'=\frac{f'g-g' f}{g^2}\)

\(\frac{\frac{d}{dx}\left(x+1\right)\left(x-1\right)-\frac{d}{dx}\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)

Mas

\(\frac{d}{dx}\left(x+1\right)=1\\ \frac{d}{dx}\left(x-1\right)=1\)

Então:

\(\frac{1\cdot \left(x-1\right)-1\cdot \left(x+1\right)}{\left(x-1\right)^2}\)

\(\frac{1\cdot \left(x-1\right)-1\cdot \left(x+1\right)}{\left(x-1\right)^2}=\boxed{\quad -\frac{2}{\left(x-1\right)^2}}\)