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We are given a function \( f(x, y) \) at the point \( (6, 5) \) and its directional derivatives in two directions. The task is to compute the gradient \( \nabla f(6, 5) \).

### Step 1: Recall the formula for the directional derivative
The directional derivative of a function \( f(x, y) \) in the direction of a unit vector \( \mathbf{u} = \langle u_1, u_2 \rangle \) is given by:
\[
D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}
\]
where \( \nabla f = \langle f_x, f_y \rangle \) is the gradient of \( f \) and \( \mathbf{u} \) is the unit vector in the given direction.

### Step 2: Extract information from the problem
We are given two directional derivatives:
1. The directional derivative in the direction parallel to \( \mathbf{v}_1 = (7, 5) \) is:
   \[
   D_{\mathbf{v}_1} f = \frac{6}{\sqrt{74}}
   \]
   The unit vector in this direction is:
   \[
   \mathbf{u}_1 = \frac{1}{\sqrt{74}}(7, 5)
   \]

2. The directional derivative in the direction parallel to \( \mathbf{v}_2 = (6, 6) \) is:
   \[
   D_{\mathbf{v}_2} f = \frac{4}{\sqrt{72}}
   \]
   The unit vector in this direction is:
   \[
   \mathbf{u}_2 = \frac{1}{\sqrt{72}}(6, 6)
   \]

### Step 3: Set up equations for the gradient
Using the formula \( D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} \), we can write two equations involving \( f_x \) and \( f_y \) based on the two directional derivatives:

1. For the direction \( (7, 5) \):
   \[
   \frac{6}{\sqrt{74}} = \frac{1}{\sqrt{74}}(7f_x + 5f_y)
   \]
   Simplifying:
   \[
   6 = 7f_x + 5f_y \quad \text{(Equation 1)}
   \]

2. For the direction \( (6, 6) \):
   \[
   \frac{4}{\sqrt{72}} = \frac{1}{\sqrt{72}}(6f_x + 6f_y)
   \]
   Simplifying:
   \[
   4 = 6f_x + 6f_y \quad \text{(Equation 2)}
   \]

### Step 4: Solve the system of equations
We now solve the system of equations:
1. \( 6 = 7f_x + 5f_y \)
2. \( 4 = 6f_x + 6f_y \)

- From Equation 2, we can simplify by dividing through by 6:
  \[
  \frac{4}{6} = f_x + f_y \quad \Rightarrow \quad \frac{2}{3} = f_x + f_y \quad \text{(Equation 3)}
  \]

- Now substitute \( f_y = \frac{2}{3} - f_x \) into Equation 1:
  \[
  6 = 7f_x + 5\left(\frac{2}{3} - f_x\right)
  \]
  Simplifying:
  \[
  6 = 7f_x + \frac{10}{3} - 5f_x
  \]
  Multiply through by 3 to eliminate the fraction:
  \[
  18 = 21f_x + 10 - 15f_x
  \]
  Simplifying:
  \[
  18 = 6f_x + 10 \quad \Rightarrow \quad 8 = 6f_x \quad \Rightarrow \quad f_x = \frac{4}{3}
  \]

- Now substitute \( f_x = \frac{4}{3} \) back into Equation 3:
  \[
  \frac{2}{3} = \frac{4}{3} + f_y \quad \Rightarrow \quad f_y = \frac{2}{3} - \frac{4}{3} = -\frac{2}{3}
  \]

### Step 5: Conclusion
The gradient of \( f \) at the point \( (6, 5) \) is:
\[
\nabla f(6, 5) = \left( \frac{4}{3}, -\frac{2}{3} \right)
\]

Would you like more details or have any further questions?

### Related Questions:
1. How do you calculate directional derivatives in three dimensions?
2. What is the geometric interpretation of the gradient vector?
3. How do you find the unit vector in a given direction?
4. What is the relation between the gradient and level curves of a function?
5. How do you solve a system of linear equations using substitution?

### Tip:
The gradient vector always points in the direction of the steepest ascent of a function at a given point.

Consider a function f(x, y) at the point (6, 5). At that point the function has directional derivatives: (6 / sqrt(74)) in the direction (parallel to) (7, 5), and (4 / sqrt(72)) in the direction (parallel to) (6, 6). The gradient of f at the point (6, 5) is:

Learn how to calculate the gradient of a function f(x, y) at the point (6, 5) using given directional derivatives in specified directions. This problem involves vector calculus concepts such as the gradient and directional derivatives. Suitable for students studying calculus.

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