Math / Calculus

Question5.3 - ine Fundamental ineorem ot (1 point)
Suppose that $F(x)=\int_{1}^{x} f(t) d t$ , where $f(t)=\int_{1}^{t^{4}} \frac{\sqrt{6+u^{4}}}{u} d u$
Find $F^{\prime \prime}(2)$ . $F^{\prime \prime}(2)=$ $\square$ Preview My Answers Submit Answers

Studdy Solution

STEP 1

What is this asking? We're given a function $F(x)$ defined as the integral of another function $f(t)$ , which itself is defined as an integral.
We need to find the second derivative of $F(x)$ at $x = 2$ . Watch out! Don't forget the chain rule when differentiating!
Also, remember the Fundamental Theorem of Calculus!

STEP 2

1. Find $F'(x)$
2. Find $f'(t)$
3. Find $F''(x)$
4. Calculate $F''(2)$

STEP 3

We're given $F(x) = \int_{1}^{x} f(t) \, dt$ .
By the Fundamental Theorem of Calculus, the derivative of an integral with respect to its upper limit is just the integrand evaluated at that upper limit.
So, $F'(x) = f(x)$ !

STEP 4

We have $f(t) = \int_{1}^{t^4} \frac{\sqrt{6 + u^4}}{u} \, du$ .
To find $f'(t)$ , we'll use the Fundamental Theorem of Calculus again, but with a twist!
The upper limit is $t^4$ , not just $t$ .

STEP 5

Let $g(z) = \int_{1}^{z} \frac{\sqrt{6 + u^4}}{u} \, du$ .
Then $f(t) = g(t^4)$ .
Now, we can use the chain rule: $f'(t) = g'(t^4) \cdot \frac{d}{dt}(t^4)$ .

STEP 6

By the Fundamental Theorem of Calculus, $g'(z) = \frac{\sqrt{6 + z^4}}{z}$ .
So, $g'(t^4) = \frac{\sqrt{6 + (t^4)^4}}{t^4} = \frac{\sqrt{6 + t^{16}}}{t^4}$ .

STEP 7

Also, $\frac{d}{dt}(t^4) = 4t^3$ .

STEP 8

Putting it all together: $f'(t) = \frac{\sqrt{6 + t^{16}}}{t^4} \cdot 4t^3 = \frac{4\sqrt{6 + t^{16}}}{t}$ .

STEP 9

We know $F'(x) = f(x)$ .
Therefore, $F''(x) = f'(x)$ .
Replacing $t$ with $x$ in our expression for $f'(t)$ , we get $F''(x) = \frac{4\sqrt{6 + x^{16}}}{x}$ .

STEP 10

Finally, we substitute $x = 2$ into $F''(x)$ : $F''(2) = \frac{4\sqrt{6 + 2^{16}}}{2} = \frac{4\sqrt{6 + 65536}}{2} = \frac{4\sqrt{65542}}{2} = 2\sqrt{65542}$

STEP 11

$F''(2) = 2\sqrt{65542}$ .

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Question5.3 - ine Fundamental ineorem ot (1 point)
Suppose that $F(x)=\int_{1}^{x} f(t) d t$ , where $f(t)=\int_{1}^{t^{4}} \frac{\sqrt{6+u^{4}}}{u} d u$
Find $F^{\prime \prime}(2)$ . $F^{\prime \prime}(2)=$ $\square$ Preview My Answers Submit Answers

Studdy Solution

STEP 1

STEP 2

1. Find $F'(x)$
2. Find $f'(t)$
3. Find $F''(x)$
4. Calculate $F''(2)$

STEP 3

We're given $F(x) = \int_{1}^{x} f(t) \, dt$ .
By the Fundamental Theorem of Calculus, the derivative of an integral with respect to its upper limit is just the integrand evaluated at that upper limit.
So, $F'(x) = f(x)$ !

STEP 4

We have $f(t) = \int_{1}^{t^4} \frac{\sqrt{6 + u^4}}{u} \, du$ .
To find $f'(t)$ , we'll use the Fundamental Theorem of Calculus again, but with a twist!
The upper limit is $t^4$ , not just $t$ .

STEP 5

Let $g(z) = \int_{1}^{z} \frac{\sqrt{6 + u^4}}{u} \, du$ .
Then $f(t) = g(t^4)$ .
Now, we can use the chain rule: $f'(t) = g'(t^4) \cdot \frac{d}{dt}(t^4)$ .

STEP 6

By the Fundamental Theorem of Calculus, $g'(z) = \frac{\sqrt{6 + z^4}}{z}$ .
So, $g'(t^4) = \frac{\sqrt{6 + (t^4)^4}}{t^4} = \frac{\sqrt{6 + t^{16}}}{t^4}$ .

STEP 7

Also, $\frac{d}{dt}(t^4) = 4t^3$ .

STEP 8

Putting it all together: $f'(t) = \frac{\sqrt{6 + t^{16}}}{t^4} \cdot 4t^3 = \frac{4\sqrt{6 + t^{16}}}{t}$ .

STEP 9

We know $F'(x) = f(x)$ .
Therefore, $F''(x) = f'(x)$ .
Replacing $t$ with $x$ in our expression for $f'(t)$ , we get $F''(x) = \frac{4\sqrt{6 + x^{16}}}{x}$ .

STEP 10

Finally, we substitute $x = 2$ into $F''(x)$ : $F''(2) = \frac{4\sqrt{6 + 2^{16}}}{2} = \frac{4\sqrt{6 + 65536}}{2} = \frac{4\sqrt{65542}}{2} = 2\sqrt{65542}$

STEP 11

$F''(2) = 2\sqrt{65542}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find F′(x)F'(x)F′(x)2. Find f′(t)f'(t)f′(t)3. Find F′′(x)F''(x)F′′(x)4. Calculate F′′(2)F''(2)F′′(2)

STEP 3

We're given F(x)=∫1xf(t) dtF(x) = \int_{1}^{x} f(t) \, dtF(x)=∫1x​f(t)dt.By the **Fundamental Theorem of Calculus**, the derivative of an integral with respect to its upper limit is just the integrand evaluated at that upper limit.So, F′(x)=f(x)F'(x) = f(x)F′(x)=f(x)!

STEP 4

We have f(t)=∫1t46+u4u duf(t) = \int_{1}^{t^4} \frac{\sqrt{6 + u^4}}{u} \, duf(t)=∫1t4​u6+u4​​du.To find f′(t)f'(t)f′(t), we'll use the **Fundamental Theorem of Calculus** again, but with a twist!The upper limit is t4t^4t4, not just ttt.

STEP 5

Let g(z)=∫1z6+u4u dug(z) = \int_{1}^{z} \frac{\sqrt{6 + u^4}}{u} \, dug(z)=∫1z​u6+u4​​du.Then f(t)=g(t4)f(t) = g(t^4)f(t)=g(t4).Now, we can use the **chain rule**: f′(t)=g′(t4)⋅ddt(t4)f'(t) = g'(t^4) \cdot \frac{d}{dt}(t^4)f′(t)=g′(t4)⋅dtd​(t4).

STEP 6

By the **Fundamental Theorem of Calculus**, g′(z)=6+z4zg'(z) = \frac{\sqrt{6 + z^4}}{z}g′(z)=z6+z4​​.So, g′(t4)=6+(t4)4t4=6+t16t4g'(t^4) = \frac{\sqrt{6 + (t^4)^4}}{t^4} = \frac{\sqrt{6 + t^{16}}}{t^4}g′(t4)=t46+(t4)4​​=t46+t16​​.

STEP 7

Also, ddt(t4)=4t3\frac{d}{dt}(t^4) = 4t^3dtd​(t4)=4t3.

STEP 8

Putting it all together: f′(t)=6+t16t4⋅4t3=46+t16tf'(t) = \frac{\sqrt{6 + t^{16}}}{t^4} \cdot 4t^3 = \frac{4\sqrt{6 + t^{16}}}{t}f′(t)=t46+t16​​⋅4t3=t46+t16​​.

STEP 9

We know F′(x)=f(x)F'(x) = f(x)F′(x)=f(x).Therefore, F′′(x)=f′(x)F''(x) = f'(x)F′′(x)=f′(x).Replacing ttt with xxx in our expression for f′(t)f'(t)f′(t), we get F′′(x)=46+x16xF''(x) = \frac{4\sqrt{6 + x^{16}}}{x}F′′(x)=x46+x16​​.

STEP 10

Finally, we substitute x=2x = 2x=2 into F′′(x)F''(x)F′′(x): F′′(2)=46+2162=46+655362=4655422=265542F''(2) = \frac{4\sqrt{6 + 2^{16}}}{2} = \frac{4\sqrt{6 + 65536}}{2} = \frac{4\sqrt{65542}}{2} = 2\sqrt{65542}F′′(2)=246+216​​=246+65536​​=2465542​​=265542​

STEP 11

F′′(2)=265542F''(2) = 2\sqrt{65542}F′′(2)=265542​.

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1. Find $F'(x)$
2. Find $f'(t)$
3. Find $F''(x)$
4. Calculate $F''(2)$

We're given $F(x) = \int_{1}^{x} f(t) \, dt$ .
By the Fundamental Theorem of Calculus, the derivative of an integral with respect to its upper limit is just the integrand evaluated at that upper limit.
So, $F'(x) = f(x)$ !

We have $f(t) = \int_{1}^{t^4} \frac{\sqrt{6 + u^4}}{u} \, du$ .
To find $f'(t)$ , we'll use the Fundamental Theorem of Calculus again, but with a twist!
The upper limit is $t^4$ , not just $t$ .

Let $g(z) = \int_{1}^{z} \frac{\sqrt{6 + u^4}}{u} \, du$ .
Then $f(t) = g(t^4)$ .
Now, we can use the chain rule: $f'(t) = g'(t^4) \cdot \frac{d}{dt}(t^4)$ .

By the Fundamental Theorem of Calculus, $g'(z) = \frac{\sqrt{6 + z^4}}{z}$ .
So, $g'(t^4) = \frac{\sqrt{6 + (t^4)^4}}{t^4} = \frac{\sqrt{6 + t^{16}}}{t^4}$ .

Also, $\frac{d}{dt}(t^4) = 4t^3$ .

Putting it all together: $f'(t) = \frac{\sqrt{6 + t^{16}}}{t^4} \cdot 4t^3 = \frac{4\sqrt{6 + t^{16}}}{t}$ .

We know $F'(x) = f(x)$ .
Therefore, $F''(x) = f'(x)$ .
Replacing $t$ with $x$ in our expression for $f'(t)$ , we get $F''(x) = \frac{4\sqrt{6 + x^{16}}}{x}$ .

Finally, we substitute $x = 2$ into $F''(x)$ : $F''(2) = \frac{4\sqrt{6 + 2^{16}}}{2} = \frac{4\sqrt{6 + 65536}}{2} = \frac{4\sqrt{65542}}{2} = 2\sqrt{65542}$

$F''(2) = 2\sqrt{65542}$ .