`f(x) = int_(x)^(x^(2))(e^(t))/(t)dt`, then `f'(t)` is equal to : (a) 0 (b) `2e` (c) `2e^2 -2` (d) `e^2-e`

Options

A. 0
B. 2e
C. `2e^(2)-2`
D. `e^(2)-e`

Answer

To solve the problem, we need to find the derivative \( f'(t) \) of the function defined by the integral:

\[ f(x) = \int_{x}^{x^2} \frac{e^t}{t} \, dt \]

### Step 1: Identify the Function and Its Limits
The function \( f(x) \) is defined as an integral from \( x \) to \( x^2 \) of the function \( g(t) = \frac{e^t}{t} \).

### Step 2: Apply the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, if we have a function defined as an integral with variable limits, the derivative can be computed using:

\[ f'(x) = g(b) \cdot \frac{db}{dx} - g(a) \cdot \frac{da}{dx} \]

where \( a = x \) and \( b = x^2 \).

### Step 3: Compute the Derivatives of the Limits
- The upper limit \( b = x^2 \) gives \( \frac{db}{dx} = 2x \).
- The lower limit \( a = x \) gives \( \frac{da}{dx} = 1 \).

### Step 4: Substitute into the Derivative Formula
Now substituting into the formula:

\[ f'(x) = g(x^2) \cdot 2x - g(x) \cdot 1 \]

### Step 5: Substitute the Function \( g(t) \)
Substituting \( g(t) = \frac{e^t}{t} \):

\[ f'(x) = \frac{e^{x^2}}{x^2} \cdot 2x - \frac{e^x}{x} \]

### Step 6: Simplify the Expression
Now simplify the expression:

\[ f'(x) = \frac{2x e^{x^2}}{x^2} - \frac{e^x}{x} = \frac{2e^{x^2}}{x} - \frac{e^x}{x} \]

### Step 7: Combine the Terms
Combining the terms gives:

\[ f'(x) = \frac{2e^{x^2} - e^x}{x} \]

### Step 8: Evaluate \( f'(t) \)
To find \( f'(t) \), replace \( x \) with \( t \):

\[ f'(t) = \frac{2e^{t^2} - e^t}{t} \]

### Final Result
Thus, the derivative \( f'(t) \) is:

\[ f'(t) = 2e^{t^2} - e^t \]