The solution should be the derivative of e ( f ( x ) ) 2 {\displaystyle e^{(f(x))^{2}}} evaluated at x= 1, . Thus applying chain rule to it, e ( f ( 1 ) ) 2 2 f ( 1 ) f ′ ( 1 ) = 0 {\displaystyle e^{(f(1))^{2}}2f(1)f'(1)=0}
evaluated at x= 1, . Thus applying chain rule to it,
