The point in the interval `[0,2pi]` where `f(x) =e^(x) sinx` has maximum slope is

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `pi`
D. `(3 pi)/(4)`

Explanation

To find the point in the interval \([0, 2\pi]\) where the function \(f(x) = e^x \sin x\) has the maximum slope, we need to follow these steps:

### Step 1: Find the first derivative \(f'(x)\)

Using the product rule for differentiation, we have:

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

Calculating the derivatives:

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

Thus, we can factor out \(e^x\):

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

### Step 2: Find the critical points of \(f'(x)\)

To find where the slope is maximum, we need to find the critical points of \(f'(x)\). This is done by setting \(f'(x) = 0\):

\[ e^x (\sin x + \cos x) = 0 \]

Since \(e^x\) is never zero, we can focus on:

\[ \sin x + \cos x = 0 \]

This simplifies to:

\[ \sin x = -\cos x \]

Dividing both sides by \(\cos x\) (assuming \(\cos x \neq 0\)) gives:

\[ \tan x = -1 \]

### Step 3: Solve for \(x\)

The solutions to \(\tan x = -1\) in the interval \([0, 2\pi]\) are:

\[ x = \frac{3\pi}{4}, \frac{7\pi}{4} \]

### Step 4: Find the second derivative \(f''(x)\)

To determine whether these points correspond to a maximum or minimum slope, we need to find the second derivative \(f''(x)\):

Using the product rule again on \(f'(x)\):

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

Applying the product rule:

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

Combining the terms:

\[ f''(x) = e^x (2\cos x) \]

### Step 5: Evaluate \(f''(x)\) at critical points

1. For \(x = \frac{3\pi}{4}\):

\[ f''\left(\frac{3\pi}{4}\right) = e^{\frac{3\pi}{4}} (2 \cos(\frac{3\pi}{4})) = e^{\frac{3\pi}{4}} (2 \cdot -\frac{1}{\sqrt{2}}) = -\sqrt{2} e^{\frac{3\pi}{4}} < 0 \]

This indicates a maximum slope at \(x = \frac{3\pi}{4}\).

2. For \(x = \frac{7\pi}{4}\):

\[ f''\left(\frac{7\pi}{4}\right) = e^{\frac{7\pi}{4}} (2 \cos(\frac{7\pi}{4})) = e^{\frac{7\pi}{4}} (2 \cdot \frac{1}{\sqrt{2}}) = \sqrt{2} e^{\frac{7\pi}{4}} > 0 \]

This indicates a minimum slope at \(x = \frac{7\pi}{4}\).

### Conclusion

The point in the interval \([0, 2\pi]\) where \(f(x) = e^x \sin x\) has the maximum slope is:

\[ \boxed{\frac{3\pi}{4}} \]