The directional derivative of the function f(x, y) = x² + y² along a line directed from (0, 0) to (1, 1), evaluated at the point x = 1, y = 1 is

The correct option is 2√2.

To find the directional derivative of the function $f(x,y)=x^2+y^2$ along a line directed from $(0,0)$ to $(1,1)$ evaluated at $(1,1)$, we follow a step-by-step mathematical derivation:

Step 1: Compute the gradient of the function $f(x,y)$.
The gradient vector $\nabla f$ is given by the partial derivatives with respect to $x$ and $y$:
$\nabla f=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}$
Calculating the partial derivatives:
$\frac{\partial f}{\partial x}=2x$
$\frac{\partial f}{\partial y}=2y$
Thus, the gradient vector is:
$\nabla f=2x\hat{i}+2y\hat{j}$

Step 2: Evaluate the gradient at the point $(1,1)$.
Substituting $x=1$ and $y=1$ into the gradient expression:
$\nabla f{|}_{(1,1)}=2\hat{i}+2\hat{j}$

Step 3: Determine the unit vector in the direction of the line.
The line is directed from the point $A(0,0)$ to $B(1,1)$. The vector representing this direction is:
$\overrightarrow{v}=(1-0)\hat{i}+(1-0)\hat{j}=\hat{i}+\stackrel{}{j}$
The magnitude of this vector is:
$\left|\overrightarrow{v}\right|=\sqrt{1^2+1^2}=\sqrt{2}$
The unit vector $\hat{u}$ in this direction is obtained by dividing the vector by its magnitude:
$\hat{u}=\frac{\overrightarrow{v}}{\left|\overrightarrow{v}\right|}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$

Step 4: Calculate the directional derivative.
The directional derivative $D_{u}f$ of the function is the dot product of the gradient vector and the unit direction vector:
$D_{u}f=\nabla f·\hat{u}$
Substituting our values:
$D_{u}f=(2\hat{i}+2\hat{j})·(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})$
$D_{u}f=2·\frac{1}{\sqrt{2}}+2·\frac{1}{\sqrt{2}}$
$D_{u}f=\frac{4}{\sqrt{2}}=2\sqrt{2}$

Thus, the directional derivative of the given function evaluated at $(1,1)$ is indeed 2√2.