Question

f(x)=x+10-11/x
f'(x)= ?

Answer 1

[tex](x+10- \frac{11}{x})' [/tex]
[tex](x)'+(10)' - \frac{(11)'*x-11*(x)'}{x^2} [/tex]
$1+ \frac{11}{x^2} = \frac{x^2+11}{x^2}$