Question

Qual é a equação da reta tangente à curva $y = \frac{1}{(x-1)^{2} }$ no ponto (0,1) ?

Answer 1

$y=\frac{1}{(x-1)^2}=(x-1)^{-2}\\\\\frac{d}{dx}[(x-1)^{-2}]=-2(x-1)^{-3}=\frac{-2}{(x-1)^3}\\\\f'(0)=\frac{-2}{(-1)^3}=2\\\\y-1=2(x-0)\\\\y=2x+1$