If `int_(x^(2))^(x^(4)) sin sqrt(t) dt,` f'(x) equals

Options

A. `sin x^(2) -sin x`
B. `4x^(3) sin x^(2)-2 sin x`
C. `x^(4) sin x^(2)-x sin x`
D. none of these

Answer

To solve the problem, we need to find the derivative of the function defined by the definite integral:

\[ f(x) = \int_{x^2}^{x^4} \sin(\sqrt{t}) \, dt \]

We will apply the Leibniz rule for differentiation under the integral sign. According to this rule, if we have:

\[ f(x) = \int_{a(x)}^{b(x)} g(t) \, dt \]

then the derivative \( f'(x) \) is given by:

\[ f'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x) \]

where \( g(t) = \sin(\sqrt{t}) \), \( a(x) = x^2 \), and \( b(x) = x^4 \).

### Step 1: Identify \( g(t) \), \( a(x) \), and \( b(x) \)
- \( g(t) = \sin(\sqrt{t}) \)
- \( a(x) = x^2 \)
- \( b(x) = x^4 \)

### Step 2: Compute \( b'(x) \) and \( a'(x) \)
- \( b'(x) = \frac{d}{dx}(x^4) = 4x^3 \)
- \( a'(x) = \frac{d}{dx}(x^2) = 2x \)

### Step 3: Evaluate \( g(b(x)) \) and \( g(a(x)) \)
- \( g(b(x)) = g(x^4) = \sin(\sqrt{x^4}) = \sin(x^2) \)
- \( g(a(x)) = g(x^2) = \sin(\sqrt{x^2}) = \sin(x) \)

### Step 4: Apply the Leibniz rule
Now we can substitute these into the Leibniz rule formula:

\[ f'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x) \]

Substituting the values we computed:

\[ f'(x) = \sin(x^2) \cdot (4x^3) - \sin(x) \cdot (2x) \]

### Step 5: Simplify the expression
Thus, we can write:

\[ f'(x) = 4x^3 \sin(x^2) - 2x \sin(x) \]

This is the final expression for \( f'(x) \).

### Final Answer
\[ f'(x) = 4x^3 \sin(x^2) - 2x \sin(x) \]