IfF(x) = f(xf(xf(x))),wheref(1) = 5, f(5) = 6, f '(1) = 2, f '(5) = 3,andf '(6) = 4,findF '(1).

Answer:F'(1)=108Step-by-step explanation:F(x) = f(xf(xf(x)))Lets takeg(x)= x f(x)F(x) = f(xf(g(x)))h(x)=f(g(x))F(x) = f(xh(x))m(x)= xh(x)F(x) = f(xh(x)) = f(m(x))F'(x)= f'(m(x)) m'(x)F'(1)= f'(m(1)) m'(1)m(x)= x h(x)m'(x)= h(x)+x h'(x)h(x)=f(g(x))h'(x)=f'(g(x)) g'(x)g(x)= x f(x)g'(x)=  f(x)+x f'(x)F'(1)= f'(m(1)) m'(1)m'(1)= h(1)+1 h'(1)h'(1)=f'(g(1)) g'(1)g'(1)=  f(1)+1 f'(1)Now by putting the valuesg'(1)=  f(1)+1 f'(1) = 5 + 1 x 2 = 7h'(1)=f'(g(1)) g'(1)g(x)= x f(x)   , g(1)= 1 f(1)  = 5h'(1)=f'(5) g'(1) = 3 x 7 =21m'(1)= h(1)+1 h'(1) h(x)=f(g(x))   , h(1) = f(g(1)) = f(5) = 6m'(1)= h(1)+1 h'(1) m'(1)=6+1 x 21m'(1)=27F'(1)= f'(m(1)) m'(1)m(x)= x h(x)  ,m(1)= 1 h(1) = 6F'(1)= f'(6) m'(1)F'(1)= f'(6) m'(1) = 4 x 27 =108F'(1)=108