$(a)$ Compute -
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \bigg[ \int_{0}^{e^x} \log ( t ) \cos^4 ( t ) \mathrm{d}t \bigg]
\end{align*}
$(b)$ For $x > 0 $ define $F ( x ) = \int_{1}^{x} t \log ( t ) \mathrm{d}t . $\\
\\$1.$ Determine the open interval(s) (if any) where $F ( x )$ is decreasing and all the open interval(s) (if any) where $F ( x )$ is increasing.\\
\\$2.$ Determine all the local minima of $F ( x )$ (if any) and all the local maxima of $F ( x )$ (if any) $.$
### Part (a)
We need to compute the derivative of the following integral with respect to \( x \):
\[
\frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{0}^{e^x} \log(t) \cos^4(t) \, \mathrm{d}t \right]
\]

To solve this, we use the Leibniz rule for differentiation under the integral sign, which is a consequence of the Fundamental Theorem of Calculus. The rule states:
\[
\frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{a(x)}^{b(x)} f(t, x) \, \mathrm{d}t \right] = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, \mathrm{d}t
\]

In our case, \( a(x) = 0 \), \( b(x) = e^x \), and \( f(t) = \log(t) \cos^4(t) \). Since \( f(t) \) does not explicitly depend on \( x \), the partial derivative \(\frac{\partial}{\partial x} f(t, x) = 0 \). Therefore, the formula simplifies to:
\[
\frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{0}^{e^x} \log(t) \cos^4(t) \, \mathrm{d}t \right] = \log(e^x) \cos^4(e^x) \cdot \frac{\mathrm{d}}{\mathrm{d}x} (e^x)
\]

Since \(\log(e^x) = x\) and \(\frac{\mathrm{d}}{\mathrm{d}x} (e^x) = e^x\), we get:
\[
\frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{0}^{e^x} \log(t) \cos^4(t) \, \mathrm{d}t \right] = x \cos^4(e^x) \cdot e^x
\]

Thus, the derivative is:
\[
\boxed{e^x x \cos^4(e^x)}
\]

### Part (b)
#### 1. Determine the open interval(s) where \( F(x) \) is decreasing and increasing.

Given:
\[
F(x) = \int_{1}^{x} t \log(t) \, \mathrm{d}t
\]

To determine where \( F(x) \) is increasing or decreasing, we first find its derivative:
\[
F'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left[ \int_{1}^{x} t \log(t) \, \mathrm{d}t \right]
\]

By the Fundamental Theorem of Calculus, we have:
\[
F'(x) = x \log(x)
\]

To determine where \( F(x) \) is increasing or decreasing, we analyze the sign of \( F'(x) \):
- \( F'(x) = x \log(x) \)

For \( x > 0 \):
- \( \log(x) < 0 \) when \( 0 < x < 1 \)
- \( \log(x) > 0 \) when \( x > 1 \)

Thus:
- \( F'(x) < 0 \) for \( 0 < x < 1 \) (decreasing)
- \( F'(x) > 0 \) for \( x > 1 \) (increasing)

#### 2. Determine all the local minima and maxima of \( F(x) \).

To find the local minima and maxima, we need to find the critical points where \( F'(x) = 0 \):
\[
x \log(x) = 0
\]

This equation is satisfied when \( x = 1 \). To determine whether this point is a local minimum or maximum, we examine the second derivative \( F''(x) \):
\[
F''(x) = \frac{\mathrm{d}}{\mathrm{d}x} (x \log(x)) = \log(x) + 1
\]

Evaluating at \( x = 1 \):
\[
F''(1) = \log(1) + 1 = 0 + 1 = 1
\]

Since \( F''(1) > 0 \), \( x = 1 \) is a local minimum.

Conclusion:
- \( F(x) \) is decreasing on \( (0, 1) \)
- \( F(x) \) is increasing on \( (1, \infty) \)
- \( F(x) \) has a local minimum at \( x = 1 \)