Iff(x) = f(xf(xf((1) = 5, f(5) = 6, f '(1) = 2, f '(5) = 3,andf '(6) = 4,findf '(1).

F'(1)=108

Step-by-step 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

i need a picture

step-by-step explanation:

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )