Answer:
F'(1)=108
Explanation:
F(x) = f(xf(xf(x)))
Lets take
g(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 values
g'(1)= f(1)+1 f'(1) = 5 + 1 x 2 = 7
h'(1)=f'(g(1)) g'(1)
g(x)= x f(x) , g(1)= 1 f(1) = 5
h'(1)=f'(5) g'(1) = 3 x 7 =21
m'(1)= h(1)+1 h'(1)
h(x)=f(g(x)) , h(1) = f(g(1)) = f(5) = 6
m'(1)= h(1)+1 h'(1)
m'(1)=6+1 x 21
m'(1)=27
F'(1)= f'(m(1)) m'(1)
m(x)= x h(x) ,m(1)= 1 h(1) = 6
F'(1)= f'(6) m'(1)
F'(1)= f'(6) m'(1) = 4 x 27 =108
F'(1)=108