If F(x) = f(xf(xf(x))), where f(1) = 5, f(5) = 6, f '(1) = 3, f '(5) = 4, and f '(6) = 5, find F '(1).

Question

If F(x) = f(xf(xf(x))), where  f(1) = 5, f(5) = 6, f '(1) = 3, f '(5) = 4, and f '(6) = 5, find  F '(1).

Answer 1

F(x) = f(xf(xf(x)))

 

F'(x) = f'(xf(xf(x)))(xf(xf(x)))'   [CHAIN RULE]

 

       = f'(xf(xf(x)))[(1)f(xf(x))+xf'(xf(x))(xf(x))']    [PRODUCT RULE]

 

       = f'(xf(xf(x)))[f(xf(x))+xf'(xf(x)){f(x)+xf'(x)}]    [PRODUCT RULE]

 

F'(1) = f'(f(f(1))[f(f(1))+f'(f(1){f(1)+f'(1)}

 

       = f'(f(5))[f(5)+f'(5){5+3}

 

       = f'(6)[6+4(8)]

 

       = 5(38)

 

       = 190

 

 

Answer 2

Someone had a twisted sense of humor to create this problem. So many parentheses!

 

The idea is to use the chain rule to find the derivative of f{xf[xf(x)]}.

 

Applying the chain rule means multiplying by the derivative of the"inside" function. This will require using the product rule a couple times.

 

Searching for a technique to make this a little easier to handle. Consider the inner most f(x) represented by "y".

 

Something like this:

 

F(x) = f[xf(xy)].

 

F'(x) = f'[xf(xy)][xf'(xy)(xy'+y)+f(xy)]

 

y(1) = 5 and y'(1) = 3.

 

F'(1) = f'([1f(5)][1f'(5)(3+5) + f(5)]

 

= f'(6)[4(8)+6]

 

= 5(38)

 

= 190