Solved on Dec 09, 2023

Evaluate the derivative ddx[(ff)(x)]\frac{d}{d x}[(f \circ f)(x)] at x=1x=1 given the provided table of f(x)f(x) and f(x)f'(x) values.

STEP 1

Assumptions
1. We have a function ff and its derivative ff' with given values at specific points.
2. We need to evaluate the derivative of the composition of ff with itself, denoted as (ff)(x)(f \circ f)(x).
3. We are specifically interested in evaluating the derivative at x=1x=1.

STEP 2

To find the derivative of the composition of two functions, we use the chain rule. The chain rule states that if we have two functions gg and hh, then the derivative of their composition (gh)(x)(g \circ h)(x) is given by:
ddx(gh)(x)=g(h(x))h(x)\frac{d}{dx}(g \circ h)(x) = g'(h(x)) \cdot h'(x)

STEP 3

In our case, we have g(x)=f(x)g(x) = f(x) and h(x)=f(x)h(x) = f(x), so the chain rule gives us:
ddx(ff)(x)=f(f(x))f(x)\frac{d}{dx}(f \circ f)(x) = f'(f(x)) \cdot f'(x)

STEP 4

Now we need to evaluate this expression at x=1x=1. First, we find f(1)f(1) and f(1)f'(1) from the given table.

STEP 5

From the table, we see that f(1)=0f(1) = 0 and f(1)=4f'(1) = 4.

STEP 6

Next, we need to find f(f(1))f'(f(1)). Since f(1)=0f(1) = 0, we need to find f(0)f'(0).

STEP 7

From the table, we see that f(0)=1f'(0) = -1.

STEP 8

Now we have all the values we need to apply the chain rule. We can substitute f(f(1))f'(f(1)) with f(0)f'(0) and f(1)f'(1) with its value.

STEP 9

Substitute the values into the chain rule expression:
ddx(ff)(1)=f(f(1))f(1)\frac{d}{dx}(f \circ f)(1) = f'(f(1)) \cdot f'(1)
ddx(ff)(1)=f(0)f(1)\frac{d}{dx}(f \circ f)(1) = f'(0) \cdot f'(1)

STEP 10

Now plug in the values we found:
ddx(ff)(1)=(1)4\frac{d}{dx}(f \circ f)(1) = (-1) \cdot 4

STEP 11

Calculate the result:
ddx(ff)(1)=4\frac{d}{dx}(f \circ f)(1) = -4
The derivative of (ff)(x)(f \circ f)(x) at x=1x=1 is 4-4.

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