Solved on Dec 09, 2023

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

STEP 1

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

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 $g$ and $h$ , then the derivative of their composition $(g \circ h)(x)$ is given by:
$\frac{d}{dx}(g \circ h)(x) = g'(h(x)) \cdot h'(x)$

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

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

STEP 1

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

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 $g$ and $h$ , then the derivative of their composition $(g \circ h)(x)$ is given by:
$\frac{d}{dx}(g \circ h)(x) = g'(h(x)) \cdot h'(x)$

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

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

STEP 1

STEP 2

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

From the table, we see that f′(0)=−1f'(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))f′(f(1)) with f′(0)f'(0)f′(0) and f′(1)f'(1)f′(1) with its value.

STEP 9

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

STEP 10

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

STEP 11

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

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

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

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

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

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

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

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

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

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

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