Math Snap

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

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