`(d)/(dx)(int_(f(x))^(g(x)) phi(t)dt)` is equal to

Options

A. `phi(g(x))-phi(f(x))`
B. `(1)/(2)[phi(g(x))]^(2)-(1)/(2)[phi(f(x))]^(2)`
C. `g'(x)phi(g(x))-f'(x)phi(f(x))`
D. `phi(g(x))g'(x)0-phi'(f(x))f'(x)`

Answer

To solve the problem \(\frac{d}{dx}\left(\int_{f(x)}^{g(x)} \phi(t) \, dt\right)\), we can use the Leibniz rule for differentiating under the integral sign. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Integral**: We have the integral \(\int_{f(x)}^{g(x)} \phi(t) \, dt\).

2. **Apply Leibniz Rule**: According to the Leibniz rule, the derivative of an integral with variable limits can be expressed as:
\[ \frac{d}{dx}\left(\int_{a(x)}^{b(x)} \phi(t) \, dt\right) = \phi(b(x)) \cdot b'(x) - \phi(a(x)) \cdot a'(x) \]
where \(a(x) = f(x)\) and \(b(x) = g(x)\).

3. **Differentiate the Limits**: In our case, we differentiate the upper limit \(g(x)\) and the lower limit \(f(x)\):
\[ \frac{d}{dx}\left(\int_{f(x)}^{g(x)} \phi(t) \, dt\right) = \phi(g(x)) \cdot g'(x) - \phi(f(x)) \cdot f'(x) \]

4. **Substitute the Function**: Here, \(\phi(g(x))\) is the value of the integrand at the upper limit and \(\phi(f(x))\) is the value at the lower limit.

5. **Final Expression**: Thus, we can write the final result as:
\[ \frac{d}{dx}\left(\int_{f(x)}^{g(x)} \phi(t) \, dt\right) = g'(x) \cdot \phi(g(x)) - f'(x) \cdot \phi(f(x)) \]

### Conclusion:
The expression \(\frac{d}{dx}\left(\int_{f(x)}^{g(x)} \phi(t) \, dt\right)\) is equal to:
\[ g'(x) \cdot \phi(g(x)) - f'(x) \cdot \phi(f(x)) \]