Posted by vicki on .
if F(x)=f(xf(xf(x))), where f(1)=2, f(2)=3, f'(1)=4, f'(2)=5, and f'(3)=6, find F'(1).

calculus 
Steve,
Here goes:
F(x) = f(xf(xf(x)))
F'(x) = f'(xf(xf(x))) * (xf(xf(x)))'
= f'(1*f(1*f(1))) * (xf(xf(x)))'
= f'(f(2)) * (xf(xf(x)))'
= f'(3) * (xf(xf(x)))'
= 6 * (xf(xf(x)))'
= 6 * (f(xf(x)) + xf'(xf(x)))
= 6 * (f(2) + xf'(2))
= 6 * (3 + 5)
= 6*8
= 48
Any other takers? Did I miss a step somewhere? 
calculus 
Mgraph,
F(x) = f(xf(xf(x)))
F'(x) = f'(xf(xf(x))) d/dx xf(xf(x))
F'(x) = f'(xf(xf(x))) [ f(xf(x)) + xf'(xf(x)) d/dx xf(x) ]
F'(x) = f'(xf(xf(x))) [ f(xf(x)) + xf'(xf(x)) [f(x) + xf'(x)] ]
F'(1) = f'(1*f(1*f(1))) [ f(1*f(1)) + 1*f'(1*f(1)) [f(1) + 1*f'(1)] ]
F'(1) = f'(f(f(1))) [ f(f(1)) + f'(f(1)) [f(1) + f'(1)] ]
F'(1) = f'(f(2)) [ f(2) + f'(2) [2 + f'(1)] ]
F'(1) = 6 [ 3 + 5 [2 + 4] ]
F'(1) = 6 [ 3 + 5 *6 ]
F'(1) = 6 [ 3 + 30 ]
F'(1) = 6 [ 33 ]
F'(1) = 196 
calculus 
Trent,
F(x) = f(xf(xf(x)))
F'(x) = f'(xf(xf(x))) d/dx xf(xf(x))
F'(x) = f'(xf(xf(x))) [ f(xf(x)) + xf'(xf(x)) d/dx xf(x) ]
F'(x) = f'(xf(xf(x))) [ f(xf(x)) + xf'(xf(x)) [f(x) + xf'(x)] ]
F'(1) = f'(1*f(1*f(1))) [ f(1*f(1)) + 1*f'(1*f(1)) [f(1) + 1*f'(1)] ]
F'(1) = f'(f(f(1))) [ f(f(1)) + f'(f(1)) [f(1) + f'(1)] ]
F'(1) = f'(f(2)) [ f(2) + f'(2) [2 + f'(1)] ]
F'(1) = 6 [ 3 + 5 [2 + 4] ]
F'(1) = 6 [ 3 + 5 *6 ]
F'(1) = 6 [ 3 + 30 ]
F'(1) = 6 [ 33 ]
F'(1) = 196