Efficient Calculation of Partial Derivatives ∂f/∂x and ∂f/∂y
For ∂f/∂x: |
∂/∂x(2x + 2y) |
= Start with the original function |
= ∂/∂x(2x) + ∂/∂x(2y) |
= Apply the sum rule for differentiation |
= 2 ∂/∂x(x) + 0 |
= Partial differentiate each term |
= 2 × 1 + 0 |
= Evaluate the partial derivatives |
= 2 |
= Simplify |
For ∂f/∂y: |
∂/∂y(2x + 2y) |
= Start with the original function |
= ∂/∂y(2x) + ∂/∂y(2y) |
= Apply the sum rule for differentiation |
= 0 + 2 ∂/∂y(y) |
= Partial differentiate each term |
= 0 + 2 × 1 |
= Evaluate the partial derivatives |
= 2 |
= Simplify |
Detailed Calculation of Partial Derivatives ∂f/∂x and ∂f/∂y for f(x, y) = 4x² + 2y
For ∂f/∂x: |
∂/∂x(4x² + 2y) |
= Start with the original function |
= ∂/∂x(4x²) + ∂/∂x(2y) |
= Apply the sum rule for differentiation |
= 2 * 4 * x¹ |
= Apply the power rule on 4x², knowing 2y is constant with respect to x |
= 8 * x |
= Simplify, 2 * 4 = 8 |
= 8x |
= Write in simplified form |
For ∂f/∂y: |
∂/∂y(4x² + 2y) |
= Start with the original function |
= 0 + ∂/∂y(2y) |
= 4x² is constant with respect to y |
= 0 + 2 |
= Differentiate 2y with respect to y |
= 2 |
= Write in simplified form |
Detailed Calculation of Partial Derivatives ∂f/∂x and ∂f/∂y for \( f(x, y) = 4x² + 2y³ + 4 \)
For ∂f/∂x: |
∂/∂x(4x² + 2y³ + 4) |
= Start with the original function |
= ∂/∂x(4x²) + ∂/∂x(2y³) + ∂/∂x(4) |
= Apply the sum rule for differentiation |
= 2 * 4 * x¹ + 0 + 0 |
= Apply the power rule on 4x²; 2y³ and 4 are constants with respect to x |
= 8 * x |
= Simplify, 2 * 4 = 8 |
= 8x |
= Write in simplified form |
For ∂f/∂y: |
∂/∂y(4x² + 2y³ + 4) |
= Start with the original function |
= 0 + ∂/∂y(2y³) + 0 |
= 4x² and 4 are constants with respect to y |
= 0 + 3 * 2 * y² |
= Differentiate 2y³ with respect to y |
= 6 * y² |
= Simplify, 3 * 2 = 6 |
= 6y² |
= Write in simplified form |
Master the Concept of Partial Derivatives with f(x, y) = x³y – x + 4
Unlocking the Secrets of ∂f/∂x:
∂/∂x(x³y – x + 4) |
Start by writing down the function f(x, y) and indicating we want the partial derivative with respect to x. |
= ∂/∂x(x³y) + ∂/∂x(-x) + ∂/∂x(4) |
Apply the linearity property of derivatives to break down into simpler parts. |
= y * 3x² – 1 |
Perform the differentiation: the derivative of x³y with respect to x is y * 3x², and the derivative of -x is -1. |
= 3x²y – 1 |
Simplify the expression to get the final result for ∂f/∂x. |
Demystifying the Partial Derivative ∂f/∂y:
∂/∂y(x³y – x + 4) |
Start by writing down the function f(x, y) and indicating we want the partial derivative with respect to y. |
= ∂/∂y(x³y) |
Since x and 4 are constants with respect to y, their derivatives become zero, leaving us with only ∂/∂y(x³y). |
= x³ |
Perform the differentiation: the derivative of x³y with respect to y is simply x³. |
Unlock the Complexity of Partial Derivatives with f(x, y) = x/y
Dive Into the Mysteries of ∂f/∂x:
∂(x/y)/∂x |
Start by indicating we want to find the partial derivative of f(x, y) = x/y with respect to x. |
∂(x * y⁻¹)/∂x |
Express x/y as x * y⁻¹ for easier differentiation. |
y⁻¹ |
Perform the differentiation. The derivative of x with respect to x is 1. Multiply by y⁻¹ to get y⁻¹. |
1/y |
Simplify y⁻¹ to 1/y, achieving the final result for ∂f/∂x. |
Discover the Partial Derivative ∂f/∂y:
∂(x/y)/∂y |
Start by indicating we want to find the partial derivative of f(x, y) = x/y with respect to y. |
∂(x * y⁻¹)/∂y |
Express x/y as x * y⁻¹ for easier differentiation. |
-x * y⁻² |
Perform the differentiation. The derivative of y⁻¹ with respect to y is -y⁻². Multiply by x to get -x * y⁻². |
-x/y² |
Simplify -x * y⁻² to -x/y², achieving the final result for ∂f/∂y. |
Dive Deep into Partial Derivatives: A Detailed Walkthrough of \( f(x, y) = xy \)
Understanding ∂f/∂x:
∂/∂x(xy) |
Initial expression for partial derivative with respect to x. |
= y ∂/∂x(x) |
Apply the differentiation rules, treating y as a constant. |
= y × 1 |
Derivative of x with respect to x is 1. |
= y |
Simplifying, we get y. |
Understanding ∂f/∂y:
∂/∂y(xy) |
Initial expression for partial derivative with respect to y. |
= x ∂/∂y(y) |
Apply the differentiation rules, treating x as a constant. |
= x × 1 |
Derivative of y with respect to y is 1. |
= x |
Simplifying, we get x. |