If `phi(x)=int_(1//x)^(sqrtx)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

Answer

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.

### Step-by-Step Solution:

1. **Identify the Function and Limits**:
\[ \phi(x) = \int_{\frac{1}{x}}^{\sqrt{x}} \sin(t^2) \, dt \]
Here, the lower limit \( g(x) = \frac{1}{x} \) and the upper limit \( h(x) = \sqrt{x} \).

2. **Apply Leibniz Rule**:
According to Leibniz rule:
\[ \phi'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x) \]
where \( f(t) = \sin(t^2) \).

3. **Calculate \( h'(x) \) and \( g'(x) \)**:
- For \( h(x) = \sqrt{x} \):
\[ h'(x) = \frac{1}{2\sqrt{x}} \]
- For \( g(x) = \frac{1}{x} \):
\[ g'(x) = -\frac{1}{x^2} \]

4. **Evaluate \( f(h(x)) \) and \( f(g(x)) \)**:
- \( f(h(x)) = f(\sqrt{x}) = \sin((\sqrt{x})^2) = \sin(x) \)
- \( f(g(x)) = f\left(\frac{1}{x}\right) = \sin\left(\left(\frac{1}{x}\right)^2\right) = \sin\left(\frac{1}{x^2}\right) \)

5. **Substitute into the Leibniz Rule**:
Now substituting these into the Leibniz formula:
\[ \phi'(x) = \sin(x) \cdot \frac{1}{2\sqrt{x}} - \sin\left(\frac{1}{x^2}\right) \cdot \left(-\frac{1}{x^2}\right) \]
Simplifying this gives:
\[ \phi'(x) = \frac{\sin(x)}{2\sqrt{x}} + \frac{\sin\left(\frac{1}{x^2}\right)}{x^2} \]

6. **Evaluate \( \phi'(1) \)**:
Now we need to evaluate \( \phi'(1) \):
\[ \phi'(1) = \frac{\sin(1)}{2\sqrt{1}} + \frac{\sin(1)}{1^2} \]
This simplifies to:
\[ \phi'(1) = \frac{\sin(1)}{2} + \sin(1) = \frac{\sin(1)}{2} + \frac{2\sin(1)}{2} = \frac{3\sin(1)}{2} \]

### Final Answer:
\[ \phi'(1) = \frac{3}{2} \sin(1) \]