If `phi` (x)`=int_(1//x)^(sqrt(x)) sin(t^(2))dt` then `phi`' (1)is equal to

Options

A. sin 1
B. 2 sin 1
C. `(3//2)sin 1`
D. none of these

Explanation

To find \(\phi'(1)\) where \(\phi(x) = \int_{\frac{1}{x}}^{\sqrt{x}} \sin(t^2) \, dt\), we will use the Leibniz rule for differentiation under the integral sign. Let's go through the steps in detail.

### Step 1: Write down the function
We start with the function:
\[ \phi(x) = \int_{\frac{1}{x}}^{\sqrt{x}} \sin(t^2) \, dt \]

### Step 2: Apply the Leibniz rule
According to the Leibniz rule, if \(\phi(x) = \int_{a(x)}^{b(x)} f(t) \, dt\), then:
\[ \phi'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \]
In our case:
- \(a(x) = \frac{1}{x}\)
- \(b(x) = \sqrt{x}\)
- \(f(t) = \sin(t^2)\)

### Step 3: Differentiate the limits
Now we need to find the derivatives of the limits:
- \(b'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}\)
- \(a'(x) = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}\)

### Step 4: Substitute into the Leibniz rule
Now substituting into the Leibniz rule:
\[ \phi'(x) = \sin((\sqrt{x})^2) \cdot \frac{1}{2\sqrt{x}} - \sin\left(\left(\frac{1}{x}\right)^2\right) \cdot \left(-\frac{1}{x^2}\right) \]
This simplifies to:
\[ \phi'(x) = \sin(x) \cdot \frac{1}{2\sqrt{x}} + \sin\left(\frac{1}{x^2}\right) \cdot \frac{1}{x^2} \]

### Step 5: Evaluate at \(x = 1\)
Now we need to evaluate \(\phi'(1)\):
\[ \phi'(1) = \sin(1) \cdot \frac{1}{2\sqrt{1}} + \sin(1) \cdot \frac{1}{1^2} \]
This simplifies to:
\[ \phi'(1) = \sin(1) \cdot \frac{1}{2} + \sin(1) \cdot 1 = \frac{1}{2} \sin(1) + \sin(1) = \frac{3}{2} \sin(1) \]

### Final Answer
Thus, the value of \(\phi'(1)\) is:
\[ \phi'(1) = \frac{3}{2} \sin(1) \]