Respuesta :

[tex] \\f(x)=\sin\frac{x-1}{x^2+1}\\ f'(x)=\cos\frac{x-1}{x^2+1}\cdot\frac{x^2+1-(x-1)\cdot2x}{(x^2+1)^2}\\ f'(x)=\frac{x^2+1-2x^2+2x}{(x^2+1)^2}\cos\frac{x-1}{x^2+1}\\ f'(x)=\frac{-x^2+2x+1}{(x^2+1)^2}\cos\frac{x-1}{x^2+1}\\ f'(x)=-\frac{x^2-2x-1}{(x^2+1)^2}\cos\frac{x-1}{x^2+1}\\ [/tex]