Original Function: f(x,y) = x + y – xy²

We start with the given function of two variables, x and y.

1. Partial Derivative with respect to x:

∂f/∂x = 1 – y²

We find the partial derivative with respect to x by treating y as a constant. Differentiating x gives 1, and differentiating -xy² gives -y².

2. Partial Derivative with respect to y:

∂f/∂y = 1 – 2xy

We find the partial derivative with respect to y by treating x as a constant. Differentiating y gives 1, and differentiating -xy² gives -2xy.

3. Gradient of f:

∇f = ⟨1 – y², 1 – 2xy⟩

The gradient is a vector containing the partial derivatives with respect to x and y. It describes the direction and rate of fastest increase of the function.

4. Evaluate the Gradient at a Specific Point (e.g., (1,1)):

∇f(1,1) = ⟨1 – 1², 1 – 2⋅1⋅1⟩ = ⟨0, -1⟩

We substitute x = 1 and y = 1 into the gradient to find its value at the point (1,1).

5. Find the Magnitude of the Gradient at (1,1):

‖∇f(1,1)‖ = √(0² + (-1)²) = 1

The magnitude of the gradient is the square root of the sum of the squares of its components. It tells us the steepness of the function at the point (1,1).

Original Function: f(x,y) = x²y – 3x + 4y²

We start with the given function of two variables, x and y.

1. Partial Derivative with respect to x:

∂f/∂x = 2xy – 3

We find the partial derivative with respect to x by treating y as a constant. Differentiating x²y gives 2xy, and differentiating -3x gives -3.

2. Partial Derivative with respect to y:

∂f/∂y = x² + 8y

We find the partial derivative with respect to y by treating x as a constant. Differentiating x²y gives x², and differentiating 4y² gives 8y.

3. Gradient of f:

∇f = ⟨2xy – 3, x² + 8y⟩

The gradient is a vector containing the partial derivatives with respect to x and y. It describes the direction and rate of fastest increase of the function.

4. Evaluate the Gradient at a Specific Point (e.g., (2,1)):

∇f(2,1) = ⟨2⋅2⋅1 – 3, 2² + 8⋅1⟩ = ⟨1, 12⟩

We substitute x = 2 and y = 1 into the gradient to find its value at the point (2,1).

5. Find the Magnitude of the Gradient at (2,1):

‖∇f(2,1)‖ = √(1² + 12²) = √145 ≈ 12.04

The magnitude of the gradient is the square root of the sum of the squares of its components. It tells us the steepness of the function at the point (2,1).

Original Function: f(x,y) = sin(xy) + cos(x) – y²

We start with the given function of two variables, x and y, involving both sine and cosine functions.

1. Partial Derivative with respect to x:

∂f/∂x = y⋅cos(xy) – sin(x)

We find the partial derivative with respect to x by treating y as a constant. Differentiating sin(xy) gives y⋅cos(xy), and differentiating cos(x) gives -sin(x).

2. Partial Derivative with respect to y:

∂f/∂y = x⋅cos(xy) – 2y

We find the partial derivative with respect to y by treating x as a constant. Differentiating sin(xy) gives x⋅cos(xy), and differentiating -y² gives -2y.

3. Gradient of f:

∇f = ⟨y⋅cos(xy) – sin(x), x⋅cos(xy) – 2y⟩

The gradient is a vector containing the partial derivatives with respect to x and y. It describes the direction and rate of fastest increase of the function.

4. Evaluate the Gradient at a Specific Point (e.g., (π,1)):

∇f(π,1) = ⟨1⋅cos(π) + 1, π⋅cos(π) – 2⟩ = ⟨0, -π – 2⟩

We substitute x = π and y = 1 into the gradient to find its value at the point (π,1).

5. Find the Magnitude of the Gradient at (π,1):

‖∇f(π,1)‖ = √(0² + (-π – 2)²) = √(π² + 4π + 4) ≈ 4.14 + π

The magnitude of the gradient is the square root of the sum of the squares of its components. It tells us the steepness of the function at the point (π,1).

Original Function: f(x,y) = eˣ⋅ln(y) – y⋅eˣ

We start with the given function of two variables, x and y, involving both exponential and natural logarithm functions.

1. Partial Derivative with respect to x:

∂f/∂x = eˣ⋅ln(y) – y⋅eˣ

We find the partial derivative with respect to x by treating y as a constant. Differentiating eˣ⋅ln(y) gives eˣ⋅ln(y), and differentiating y⋅eˣ gives y⋅eˣ.

2. Partial Derivative with respect to y:

∂f/∂y = eˣ/y – eˣ

We find the partial derivative with respect to y by treating x as a constant. Differentiating eˣ⋅ln(y) gives eˣ/y, and differentiating y⋅eˣ gives -eˣ.

3. Gradient of f:

∇f = ⟨eˣ⋅ln(y) – y⋅eˣ, eˣ/y – eˣ⟩

The gradient is a vector containing the partial derivatives with respect to x and y. It describes the direction and rate of fastest increase of the function.

4. Evaluate the Gradient at a Specific Point (e.g., (1, e)):

∇f(1, e) = ⟨e⋅ln(e) – e⋅e, e/e – e⟩ = ⟨e – e², 1 – e⟩

We substitute x = 1 and y = e into the gradient to find its value at the point (1, e).

5. Find the Magnitude of the Gradient at (1, e):

‖∇f(1, e)‖ = √((e – e²)² + (1 – e)²) ≈ √(1 – 2e + 2e³)

The magnitude of the gradient is the square root of the sum of the squares of its components. It tells us the steepness of the function at the point (1, e).