To find the derivative of the function given by the integral,
[tex]\[
f(x) = \int_{-3}^{x^4} \sqrt{t} \cos(t^2) \, dt,
\][/tex]
we can use the Leibniz rule for differentiating an integral with variable limits of integration.
First, let us state the Leibniz rule:
If [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] are continuous and differentiable functions, and [tex]\( F(t) \)[/tex] is a continuous function, then
[tex]\[
\frac{d}{dx} \left( \int_{g(x)}^{h(x)} F(t) \, dt \right) = F(h(x)) \cdot h'(x) - F(g(x)) \cdot g'(x).
\][/tex]
Here, [tex]\( g(x) = -3 \)[/tex] and [tex]\( h(x) = x^4 \)[/tex]. The integrand [tex]\( F(t) = \sqrt{t} \cos(t^2) \)[/tex].
Applying the Leibniz rule:
1. Evaluate [tex]\( F(t) = \sqrt{t} \cos(t^2) \)[/tex] at the upper limit [tex]\( h(x) = x^4 \)[/tex]:
[tex]\[
F(x^4) = \sqrt{x^4} \cos((x^4)^2) = x^2 \cos(x^8).
\][/tex]
2. Multiply by the derivative of the upper limit [tex]\( h(x) \)[/tex]:
[tex]\[
h'(x) = \frac{d}{dx} (x^4) = 4x^3.
\][/tex]
3. Evaluate [tex]\( F(t) = \sqrt{t} \cos(t^2) \)[/tex] at the lower limit [tex]\( g(x) = -3 \)[/tex]:
[tex]\[
F(-3) = \sqrt{-3} \cos((-3)^2) = \sqrt{-3} \cos(9) = i\sqrt{3} \cos(9).
\][/tex]
(Note: Because [tex]\(\sqrt{-3}\)[/tex] introduces an imaginary component, we have replaced it with [tex]\(i\sqrt{3}\)[/tex], considering that [tex]\(i\)[/tex] is the imaginary unit).
4. Multiply by the derivative of the lower limit [tex]\( g(x) \)[/tex]:
[tex]\[
g'(x) = \frac{d}{dx} (-3) = 0.
\][/tex]
Putting it all together:
[tex]\[
f'(x) = F(x^4) \cdot h'(x) - F(-3) \cdot g'(x).
\][/tex]
[tex]\[
f'(x) = (x^2 \cos(x^8)) (4x^3) - (i\sqrt{3} \cos(9)) \cdot 0.
\][/tex]
[tex]\[
f'(x) = 4x^5 \cos(x^8).
\][/tex]
Hence, the derivative of the function [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[
f'(x) = 4x^5 \cos(x^8).
\][/tex]
