If `f(x)= int_(x^(2))^(x^(4)) sin sqrtt dt`, then f'(x) equals

Options

A. `sin x^(2) -sin x`
B. `4x^(3) sin x^(2) - 2x 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 \( f'(x) \) of the function defined by the integral:

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

We will use the Fundamental Theorem of Calculus and the Leibniz Rule for differentiation under the integral sign.

### Step 1: Apply the Leibniz Rule
The Leibniz Rule states that if \( F(x) = \int_{a(x)}^{b(x)} f(t) \, dt \), then the derivative \( F'(x) \) is given by:

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

In our case, \( a(x) = x^2 \) and \( b(x) = x^4 \), and \( f(t) = \sin(\sqrt{t}) \).

### Step 2: Compute the derivatives of the limits
Now, we need to compute the derivatives of the limits:

- \( a'(x) = \frac{d}{dx}(x^2) = 2x \)
- \( b'(x) = \frac{d}{dx}(x^4) = 4x^3 \)

### Step 3: Evaluate \( f(b(x)) \) and \( f(a(x)) \)
Next, we evaluate \( f(b(x)) \) and \( f(a(x)) \):

- \( f(b(x)) = f(x^4) = \sin(\sqrt{x^4}) = \sin(x^2) \)
- \( f(a(x)) = f(x^2) = \sin(\sqrt{x^2}) = \sin(x) \)

### Step 4: Substitute into the Leibniz Rule
Now we can substitute everything back into the Leibniz Rule:

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

Substituting the values we calculated:

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

### Final Expression
Thus, the derivative \( f'(x) \) is:

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