Find the point at which the slope of the tangent of the function `f(x)=e^xcosx` attains minima, when `x in [0,2pi]dot`

Explanation

To find the point at which the slope of the tangent of the function \( f(x) = e^x \cos x \) attains a minimum value in the interval \( x \in [0, 2\pi] \), we will follow these steps:

### Step 1: Find the first derivative of the function
The first step is to find the derivative of the function \( f(x) \).

\[ f(x) = e^x \cos x \]

Using the product rule, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(e^x) \cdot \cos x + e^x \cdot \frac{d}{dx}(\cos x) \]

Calculating the derivatives:

\[ f'(x) = e^x \cos x - e^x \sin x = e^x (\cos x - \sin x) \]

### Step 2: Define the function representing the slope of the tangent
Let \( g(x) = f'(x) = e^x (\cos x - \sin x) \). We need to find the minimum of this function \( g(x) \).

### Step 3: Find the second derivative of \( g(x) \)
To find the critical points where the slope of the tangent might attain a minimum, we need to differentiate \( g(x) \):

\[ g'(x) = \frac{d}{dx}(e^x (\cos x - \sin x)) \]

Using the product rule again:

\[ g'(x) = e^x (\cos x - \sin x) + e^x (-\sin x - \cos x) \]

Simplifying this:

\[ g'(x) = e^x (\cos x - \sin x - \sin x - \cos x) = e^x (-2\sin x) \]

### Step 4: Set the first derivative equal to zero
To find critical points, set \( g'(x) = 0 \):

\[ e^x (-2\sin x) = 0 \]

Since \( e^x \) is never zero, we have:

\[ -2\sin x = 0 \implies \sin x = 0 \]

The solutions for \( \sin x = 0 \) in the interval \( [0, 2\pi] \) are:

\[ x = 0, \pi, 2\pi \]

### Step 5: Determine the nature of critical points
Next, we need to determine whether these points are minima or maxima. We can use the second derivative test or analyze the sign of \( g'(x) \) around these points.

1. For \( x = 0 \):
- \( g'(x) \) changes from positive to negative (increasing to decreasing).
- This indicates a local maximum.

2. For \( x = \pi \):
- \( g'(x) \) changes from negative to positive (decreasing to increasing).
- This indicates a local minimum.

3. For \( x = 2\pi \):
- \( g'(x) \) changes from positive to negative (increasing to decreasing).
- This indicates a local maximum.

### Conclusion
The slope of the tangent \( g(x) \) attains its minimum value at:

\[ \boxed{x = \pi} \]