$$ \frac { d } { d x } \int _ { x } ^ { 2 } \sqrt { t ^ { 5 } + 4 } d t = \square $$

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To solve the expression

$$
\frac{d}{dx} \int_{x}^{2} \sqrt{t^{5} + 4} \, dt,
$$

we can apply the Leibniz rule for differentiation under the integral sign. According to this rule, if we have an integral of the form

$$
F(x) = \int_{a(x)}^{b(x)} f(t) \, dt,
$$

the derivative is given by:

$$
F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x).
$$

In our case:

- $$ a(x) = x $$ and $$ b(x) = 2 $$
- $$ f(t) = \sqrt{t^{5} + 4} $$

Now, we can compute the derivative:

1. **Evaluate $$ f(b(x)) $$**:
   - Since $$ b(x) = 2 $$, we have $$ f(2) = \sqrt{2^{5} + 4} = \sqrt{32 + 4} = \sqrt{36} = 6 $$.

2. **Evaluate $$ f(a(x)) $$**:
   - Since $$ a(x) = x $$, we have $$ f(x) = \sqrt{x^{5} + 4} $$.

3. **Derivatives of the limits**:
   - $$ b'(x) = 0 $$ (since $$ b(x) = 2 $$ is constant)
   - $$ a'(x) = 1 $$ (since $$ a(x) = x $$)

Putting it all together, we have:

$$
\frac{d}{dx} \int_{x}^{2} \sqrt{t^{5} + 4} \, dt = f(2) \cdot 0 - f(x) \cdot 1 = -\sqrt{x^{5} + 4}.
$$

Thus, the final result is:

$$
\frac{d}{dx} \int_{x}^{2} \sqrt{t^{5} + 4} \, dt = -\sqrt{x^{5} + 4}.
$$
$$ \frac{d}{dx} \int_{x}^{2} \sqrt{t^{5} + 4} \, dt = -\sqrt{x^{5} + 4} $$

\( \frac { d } { d x } \int _ { x } ^ { 2 } \sqrt { t ^ { 5 } + 4 } d t = \square \)

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