Math / Calculus

QuestionProblem 9. (1 point)
A table of values for $f, g, f^{\prime}$ , and $g^{\prime}$ is given below. \begin{tabular}{|c|c|c|c|c|} \hline $x$ & $f(x)$ & $g(x)$ & $f^{\prime}(x)$ & $g^{\prime}(x)$ \\ \hline 1 & 3 & 1 & 2 & 2 \\ \hline 2 & 2 & 2 & 2 & 2 \\ \hline 3 & 1 & 1 & 2 & 3 \\ \hline \end{tabular} (A) If $h(x)=f(g(x))$ , then $h^{\prime}(2)=$ $\square$ (B) If $H(x)=g(f(x))$ , then $H^{\prime}(1)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the derivatives of composite functions at specific points, using a given table of values. Watch out! Don't mix up $f$ and $g$ , and remember the chain rule: the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.

STEP 2

1. Find $h'(2)$
2. Find $H'(1)$

STEP 3

Alright, let's start with $h(x) = f(g(x))$ .
We want to find $h'(2)$ , which means we need the derivative of $h$ .

STEP 4

Time for the chain rule!
The derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$ .
So, $h'(x) = f'(g(x)) \cdot g'(x)$ .

STEP 5

Now, we plug in $x = 2$ .
We get $h'(2) = f'(g(2)) \cdot g'(2)$ .

STEP 6

Let's grab those values from the table. $g(2)$ is 2, and $g'(2)$ is 2.
Substituting these values, we have $h'(2) = f'(2) \cdot 2$ .

STEP 7

Back to the table! $f'(2)$ is 2.
So, $h'(2) = 2 \cdot 2 = 4$ .
Boom!

STEP 8

Now for $H(x) = g(f(x))$ .
We want $H'(1)$ , so we need to differentiate $H(x)$ .

STEP 9

Chain rule time again!
The derivative of $g(f(x))$ is $g'(f(x)) \cdot f'(x)$ .
Therefore, $H'(x) = g'(f(x)) \cdot f'(x)$ .

STEP 10

Let's substitute $x = 1$ .
We get $H'(1) = g'(f(1)) \cdot f'(1)$ .

STEP 11

From the table, $f(1)$ is 3 and $f'(1)$ is 2.
So, $H'(1) = g'(3) \cdot 2$ .

STEP 12

One last trip to the table! $g'(3)$ is 3.
Thus, $H'(1) = 3 \cdot 2 = 6$ .
Fantastic!

STEP 13

$h'(2) = 4$ $H'(1) = 6$

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Find $h'(2)$
2. Find $H'(1)$

STEP 3

Alright, let's start with $h(x) = f(g(x))$ .
We want to find $h'(2)$ , which means we need the derivative of $h$ .

STEP 4

Time for the chain rule!
The derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$ .
So, $h'(x) = f'(g(x)) \cdot g'(x)$ .

STEP 5

Now, we plug in $x = 2$ .
We get $h'(2) = f'(g(2)) \cdot g'(2)$ .

STEP 6

Let's grab those values from the table. $g(2)$ is 2, and $g'(2)$ is 2.
Substituting these values, we have $h'(2) = f'(2) \cdot 2$ .

STEP 7

Back to the table! $f'(2)$ is 2.
So, $h'(2) = 2 \cdot 2 = 4$ .
Boom!

STEP 8

Now for $H(x) = g(f(x))$ .
We want $H'(1)$ , so we need to differentiate $H(x)$ .

STEP 9

Chain rule time again!
The derivative of $g(f(x))$ is $g'(f(x)) \cdot f'(x)$ .
Therefore, $H'(x) = g'(f(x)) \cdot f'(x)$ .

STEP 10

Let's substitute $x = 1$ .
We get $H'(1) = g'(f(1)) \cdot f'(1)$ .

STEP 11

From the table, $f(1)$ is 3 and $f'(1)$ is 2.
So, $H'(1) = g'(3) \cdot 2$ .

STEP 12

One last trip to the table! $g'(3)$ is 3.
Thus, $H'(1) = 3 \cdot 2 = 6$ .
Fantastic!

STEP 13

$h'(2) = 4$ $H'(1) = 6$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Find h′(2)h'(2)h′(2)2. Find H′(1)H'(1)H′(1)

STEP 3

Alright, let's **start** with h(x)=f(g(x))h(x) = f(g(x))h(x)=f(g(x)).We want to find h′(2)h'(2)h′(2), which means we need the derivative of hhh.

STEP 4

Time for the **chain rule**!The derivative of f(g(x))f(g(x))f(g(x)) is f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x)f′(g(x))⋅g′(x).So, h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x)h′(x)=f′(g(x))⋅g′(x).

STEP 5

Now, we **plug in** x=2x = 2x=2.We get h′(2)=f′(g(2))⋅g′(2)h'(2) = f'(g(2)) \cdot g'(2)h′(2)=f′(g(2))⋅g′(2).

STEP 6

Let's **grab those values** from the table. g(2)g(2)g(2) is **2**, and g′(2)g'(2)g′(2) is **2**.Substituting these values, we have h′(2)=f′(2)⋅2h'(2) = f'(2) \cdot 2h′(2)=f′(2)⋅2.

STEP 7

Back to the **table**! f′(2)f'(2)f′(2) is **2**.So, h′(2)=2⋅2=4h'(2) = 2 \cdot 2 = 4h′(2)=2⋅2=4.Boom!

STEP 8

Now for H(x)=g(f(x))H(x) = g(f(x))H(x)=g(f(x)).We want H′(1)H'(1)H′(1), so we need to **differentiate** H(x)H(x)H(x).

STEP 9

**Chain rule** time again!The derivative of g(f(x))g(f(x))g(f(x)) is g′(f(x))⋅f′(x)g'(f(x)) \cdot f'(x)g′(f(x))⋅f′(x).Therefore, H′(x)=g′(f(x))⋅f′(x)H'(x) = g'(f(x)) \cdot f'(x)H′(x)=g′(f(x))⋅f′(x).

STEP 10

Let's **substitute** x=1x = 1x=1.We get H′(1)=g′(f(1))⋅f′(1)H'(1) = g'(f(1)) \cdot f'(1)H′(1)=g′(f(1))⋅f′(1).

STEP 11

From the **table**, f(1)f(1)f(1) is **3** and f′(1)f'(1)f′(1) is **2**.So, H′(1)=g′(3)⋅2H'(1) = g'(3) \cdot 2H′(1)=g′(3)⋅2.

STEP 12

One last **trip to the table**! g′(3)g'(3)g′(3) is **3**.Thus, H′(1)=3⋅2=6H'(1) = 3 \cdot 2 = 6H′(1)=3⋅2=6.Fantastic!

STEP 13

h′(2)=4h'(2) = 4h′(2)=4 H′(1)=6H'(1) = 6H′(1)=6

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Find $h'(2)$
2. Find $H'(1)$

Alright, let's start with $h(x) = f(g(x))$ .
We want to find $h'(2)$ , which means we need the derivative of $h$ .

Time for the chain rule!
The derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$ .
So, $h'(x) = f'(g(x)) \cdot g'(x)$ .

Now, we plug in $x = 2$ .
We get $h'(2) = f'(g(2)) \cdot g'(2)$ .

Let's grab those values from the table. $g(2)$ is 2, and $g'(2)$ is 2.
Substituting these values, we have $h'(2) = f'(2) \cdot 2$ .

Back to the table! $f'(2)$ is 2.
So, $h'(2) = 2 \cdot 2 = 4$ .
Boom!

Now for $H(x) = g(f(x))$ .
We want $H'(1)$ , so we need to differentiate $H(x)$ .

Chain rule time again!
The derivative of $g(f(x))$ is $g'(f(x)) \cdot f'(x)$ .
Therefore, $H'(x) = g'(f(x)) \cdot f'(x)$ .

Let's substitute $x = 1$ .
We get $H'(1) = g'(f(1)) \cdot f'(1)$ .

From the table, $f(1)$ is 3 and $f'(1)$ is 2.
So, $H'(1) = g'(3) \cdot 2$ .

One last trip to the table! $g'(3)$ is 3.
Thus, $H'(1) = 3 \cdot 2 = 6$ .
Fantastic!

$h'(2) = 4$ $H'(1) = 6$