Question

A differentiation problem generated by
http://saab.org/calculus.cgi

Answer 1

\[g(x)=\frac{x+f(x)}{x-f(x)}\]By the quotient rule, \[g'(x)=\frac{(x-f(x))(1+f'(x))-(x+f(x))(1-f'(x))}{(x-f(x))^2}\]Thus\[g'(11)=\frac{(11-f(11))(1+f'(11))-(11+f(11))(1-f'(11))}{(11-f(11))^2}\]

Answer 2

\[g'(x) = \frac{ x+f(x) }{ x-f(x) }\left( \frac{ 1+f'(x) }{ x+f(x) }-\frac{ 1-f'(x) }{ x-f(x) } \right)\]

g'(11) = 17/18

Answer 3

Solution generated by http://saab.org/calculus.cgi