The derivative of 

f(x,y)=3x2+4yf(x,y)=3x2+4y

 at 

P(3,2)P(3,2)

 in the direction of 

i^+j^i^+j^

 is 

To find the derivative of the function \( f(x, y) = 3x^2 + 4y \) at the point \( P(3, 2) \) in the direction of the vector \( \mathbf{i} + \mathbf{j} \), we follow these steps:

### 1. **Find the Gradient of \( f(x, y) \):**
The gradient of \( f(x, y) \) is a vector that consists of the partial derivatives with respect to \( x \) and \( y \):

\[
\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]

Calculate the partial derivatives:

\[
\frac{\partial f}{\partial x} = 6x, \quad \frac{\partial f}{\partial y} = 4
\]

So, the gradient is:

\[
\mathbf{\nabla} f(x, y) = (6x, 4)
\]

### 2. **Evaluate the Gradient at \( P(3, 2) \):**
Substitute \( x = 3 \) and \( y = 2 \) into the gradient:

\[
\mathbf{\nabla} f(3, 2) = (6 \times 3, 4) = (18, 4)
\]

### 3. **Find the Unit Vector in the Direction \( \mathbf{i} + \mathbf{j} \):**
The direction vector given is \( \mathbf{i} + \mathbf{j} \), which in component form is \( \mathbf{d} = (1, 1) \).

To convert this to a unit vector, we calculate its magnitude:

\[
\| \mathbf{d} \| = \sqrt{1^2 + 1^2} = \sqrt{2}
\]

The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{i} + \mathbf{j} \) is:

\[
\mathbf{u} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)
\]

### 4. **Compute the Directional Derivative:**
The directional derivative \( D_{\mathbf{u}} f \) of \( f(x, y) \) at \( P(3, 2) \) in the direction of \( \mathbf{u} \) is given by the dot product of the gradient and the unit vector:

\[
D_{\mathbf{u}} f = \mathbf{\nabla} f(3, 2) \cdot \mathbf{u} = (18, 4) \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)
\]

Calculate the dot product:

\[
D_{\mathbf{u}} f = \frac{18}{\sqrt{2}} + \frac{4}{\sqrt{2}} = \frac{22}{\sqrt{2}} = \frac{22 \sqrt{2}}{2} = 11\sqrt{2}
\]

### Final Answer:
The derivative of \( f(x, y) = 3x^2 + 4y \) at \( P(3, 2) \) in the direction of \( \mathbf{i} + \mathbf{j} \) is \( 11\sqrt{2} \).

Would you like further explanation on any of these steps, or do you have any questions?

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Here are 8 related questions:
1. How do you find the gradient of a function in three dimensions?
2. What is the physical significance of the directional derivative?
3. How does the directional derivative relate to the gradient and the angle between vectors?
4. Can you compute the directional derivative in a direction other than \( \mathbf{i} + \mathbf{j} \)?
5. How does the directional derivative change if the point \( P(x, y) \) is different?
6. What happens if the direction vector is not normalized before computing the directional derivative?
7. How do you determine the maximum rate of change of a function at a given point?
8. What is the relationship between the directional derivative and the tangent plane at a point?

**Tip:** Always remember to normalize the direction vector when calculating the directional derivative to ensure accuracy.

Learn how to calculate the derivative of the function f(x, y) = 3x^2 + 4y at the point P(3, 2) in the direction of the vector i + j. Understand the gradient, directional derivative, and vector calculus concepts. Suitable for advanced undergraduate students and learners of multivariable calculus.

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