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To find the directional derivative \( D_{\mathbf{u}} f(x, y) \) of \( f(x, y) = e^x \sin y \) at the point \( P\left(0, \frac{\pi}{3}\right) \) in the direction of the vector \( \mathbf{v} = -6\mathbf{i} + 8\mathbf{j} \), we will follow these steps:

### Step 1: Compute the Gradient of \( f(x, y) \)
The gradient of \( f(x, y) \), denoted as \( \nabla f(x, y) \), is:
\[
\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]
For \( f(x, y) = e^x \sin y \):
1. **Partial derivative with respect to \( x \):**
   \[
   \frac{\partial f}{\partial x} = e^x \sin y
   \]
2. **Partial derivative with respect to \( y \):**
   \[
   \frac{\partial f}{\partial y} = e^x \cos y
   \]
   
Thus,
\[
\nabla f(x, y) = \left( e^x \sin y, e^x \cos y \right)
\]

### Step 2: Evaluate \( \nabla f(x, y) \) at \( P\left(0, \frac{\pi}{3}\right) \)
Substitute \( x = 0 \) and \( y = \frac{\pi}{3} \) into the gradient:
1. **For \( \frac{\partial f}{\partial x} \):**
   \[
   e^0 \sin\left(\frac{\pi}{3}\right) = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}
   \]
2. **For \( \frac{\partial f}{\partial y} \):**
   \[
   e^0 \cos\left(\frac{\pi}{3}\right) = 1 \cdot \frac{1}{2} = \frac{1}{2}
   \]

So, \( \nabla f\left(0, \frac{\pi}{3}\right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).

### Step 3: Normalize the Direction Vector \( \mathbf{v} \)
The direction vector \( \mathbf{v} = -6\mathbf{i} + 8\mathbf{j} \) needs to be converted into a unit vector \( \mathbf{u} \).

1. **Magnitude of \( \mathbf{v} \):**
   \[
   \|\mathbf{v}\| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10
   \]
2. **Unit vector \( \mathbf{u} \):**
   \[
   \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{-6\mathbf{i} + 8\mathbf{j}}{10} = -0.6\mathbf{i} + 0.8\mathbf{j}
   \]

### Step 4: Compute the Directional Derivative
The directional derivative of \( f \) at \( P\left(0, \frac{\pi}{3}\right) \) in the direction of \( \mathbf{u} \) is:
\[
D_{\mathbf{u}} f(x, y) = \nabla f(x, y) \cdot \mathbf{u}
\]
At \( P\left(0, \frac{\pi}{3}\right) \):
\[
D_{\mathbf{u}} f\left(0, \frac{\pi}{3}\right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \cdot \left( -0.6, 0

Find the directional derivative for f(x, y) = e^x sin y in the direction of vector v = -6i + 8j at the point P(0, π/3).

Explore how to calculate the directional derivative of f(x, y) = e^x sin y at the point (0, π/3) in the direction of vector v = -6i + 8j. This solution covers the gradient calculation, unit vector normalization, and applying the directional derivative formula.

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