Use Chain Rule to Find Derivative: f(1) = 2, f(2) = 3, f'(1) = 4, f'(2) = 5, f'(3) = 6; g(x) = f ( x f ( x f ( x ) ) ), find g'(1)

[OkCoolUserNameDude]

I'm completely stuck on trying to make the equation for chain rule. I can't get passed making dy/dx = dy/du * du/dz * dz/dx, I have no idea what to put here.

Information:
f(1) = 2, f(2) = 3, f'(1) = 4, f'(2) = 5, and f'(3) = 6

if g(x) = f ( x f ( x f ( x ) ) ), find g'(1)

My steps:
y = f(u)
u = xf(z)
z = xf(x)

Made equation:

dy/dx = dy/du * du/dz * dz/dx

Stuck and don't know where to go past this :/
Do I try to figure out an equation for f(x)???

 

[Steven G]

We appreciate you showing your work! What does f ( x f ( x f ( x ) ) ) mean???????

 

[pka]

We appreciate you showing your work! What does f ( x f ( x f ( x ) ) ) ???????

I wish I could actually read the attachment. Here I have a very strong objection to overkill.
Jomo, to answer your question: \(\displaystyle \mathcal{D_x}[f(x\cdot f(x))]=[f'(x\cdot f(x))]\cdot[f(x)+x\cdot f'(x)]\) .
Now in the posted problem there is yet another level of composition (that is the overkill).

 

[Dr.Peterson]

Information:
f(1) = 2, f(2) = 3, f'(1) = 4, f'(2) = 5, and f'(3) = 6

if g(x) = f ( x f ( x f ( x ) ) ), find g'(1)

My steps:
y = f(u)
u = xf(z)
z = xf(x)

Made equation:

dy/dx = dy/du * du/dz * dz/dx

Stuck and don't know where to go past this :/
Do I try to figure out an equation for f(x)???

I wouldn't write that equation, which focuses your attention on the wrong things, and emphasizes the complexity. I'd just take it step by step, doing one thing that you know how to do at a time.

First, we have f(x f(x f(x))), and can think of u = x f(x f(x)), so we need to differentiate (with respect to x) f(u). Applying the chain rule to this, we have f'(u) * u'(x). The first factor can be evaluated using what we're given: f'(u) = f'(x f(x f(x))), so f'(1) = f'(1 f(1 f(1))) = ... . (We're hoping the argument will end up being something we can handle, namely 1, 2, or 3! It will be.)

Now you have to find u'(x), that is, d/dx [x f(x f(x))]. For this, you can use the product rule. Then you'll use the chain rule again, and so on.