To find the derivative of the function given by
[tex]\[ f(x) = \int_x^0 \sqrt{6 + \sec(8t)} \, dt \][/tex]
using the Fundamental Theorem of Calculus, we can follow these steps:
### Step 1: Rewrite the Integral
Firstly, note the hint provided:
[tex]\[ \int_x^0 \sqrt{6 + \sec(8t)} \, dt = -\int_0^x \sqrt{6 + \sec(8t)} \, dt. \][/tex]
This allows us to rewrite [tex]\( f(x) \)[/tex] as:
[tex]\[ f(x) = -\int_0^x \sqrt{6 + \sec(8t)} \, dt. \][/tex]
### Step 2: Apply the Fundamental Theorem of Calculus
The fundamental theorem of calculus states that if [tex]\(F(x) = \int_{a}^{x} g(t) \, dt\)[/tex], then [tex]\(F'(x) = g(x)\)[/tex].
Applying this to our function:
[tex]\[ \int_0^x \sqrt{6 + \sec(8t)} \, dt \][/tex]
Let [tex]\( G(x) = \int_0^x \sqrt{6 + \sec(8t)} \, dt \)[/tex], then by the Fundamental Theorem of Calculus:
[tex]\[ G'(x) = \sqrt{6 + \sec(8x)} \][/tex]
### Step 3: Differentiate and Apply the Negative Sign
Since:
[tex]\[ f(x) = -G(x) \][/tex]
We need to find [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = -(G'(x)) \][/tex]
Thus,
[tex]\[ f'(x) = -\sqrt{6 + \sec(8x)} \][/tex]
### Conclusion
Therefore, the derivative of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = -\sqrt{6 + \sec(8x)} \][/tex]
