I) Find the directional derivative of 𝑓(𝑥, 𝑦) = 𝑥 sin 𝑦 at (1,0) in the direction of the unit vector that make an angle of 𝜋/4 with positive 𝑥-axis.

To find the directional derivative of a function, we need to calculate the dot product of the gradient vector of the function and the unit vector in the given direction.

The gradient vector of the function f(x, y) is given by the partial derivatives of f with respect to each variable.

Let's find the partial derivatives of f(x, y):

\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x \sin y) = \sin y

\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x \sin y) = x \cos y

Now, we can find the gradient vector:

\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = \left(\sin y, x \cos y\right)

The unit vector that makes an angle of 𝜋/4 with the positive 𝑥-axis is:

\mathbf{u}=\left(\cos\frac{\pi}{4},\sin\frac{\pi}{4}\right)=\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)

Now, we can calculate the dot product of the gradient vector and the unit vector:

\nabla f\cdot\mathbf{u}=\left(\sin y,x\cos y\right)\cdot\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}\sin y+\frac{\sqrt{2}}{2}x\cos y

Substituting the coordinates of the point (1, 0) into the equation above:

\nabla f\cdot\mathbf{u}=\frac{\sqrt{2}}{2}\cdot\sin0+\frac{\sqrt{2}}{2}\cdot1\cdot\cos0=\frac{\sqrt{2}}{2}\cdot0+\frac{\sqrt{2}}{2}\cdot1\cdot1=\frac{\sqrt{2}}{2}

Answer: The directional derivative of f(x, y) = x sin y at (1, 0) in the direction of the unit vector that makes an angle of 𝜋/4 with the positive 𝑥-axis is \frac{\sqrt{2}}{2} .