Last updated on July 19th, 2025
The derivative of a multivariable function is a tool used to understand how a function changes in response to small changes in its input variables. This concept is crucial for calculating rates of change in various fields, including economics and physics. In this discussion, we will explore the derivative of a multivariable function in detail.
The derivative of a multivariable function is represented as a gradient vector, which contains all the partial derivatives of the function with respect to each of its input variables. This type of derivative indicates how the function changes along each axis in its domain and is differentiable within its domain. Key concepts include: - Partial Derivatives: These are derivatives of a function with respect to one variable while keeping other variables constant. - Gradient Vector: A vector that contains all the partial derivatives of the function. - Directional Derivative: Derivative of the function in a specific direction in the input space.
The derivative of a multivariable function f(x, y, ...) can be expressed as the gradient vector: ∇f = (∂f/∂x, ∂f/∂y, ...) This formula applies to functions with continuous partial derivatives within their domain.
We can derive the derivative of a multivariable function using different approaches. The fundamental methods include: - By First Principles: Involves taking limits of difference quotients for each partial derivative. - Using Chain Rule: Applies when functions are composed of other functions. - Using Total Differentiation: Combines all partial derivatives to understand changes in the function. By First Principles The derivative of a multivariable function can be shown using first principles by taking the limit of the difference quotient for each variable. Consider a function f(x, y). The partial derivative of f with respect to x is: ∂f/∂x = limΔx→0 [f(x + Δx, y) - f(x, y)] / Δx Similarly, the partial derivative with respect to y is: ∂f/∂y = limΔy→0 [f(x, y + Δy) - f(x, y)] / Δy Using Chain Rule For functions of the form f(g(x, y), h(x, y)), the chain rule is used to find derivatives with respect to x and y. - ∂f/∂x = (∂f/∂u)(∂u/∂x) + (∂f/∂v)(∂v/∂x) - ∂f/∂y = (∂f/∂u)(∂u/∂y) + (∂f/∂v)(∂v/∂y) Using Total Differentiation Total differentiation involves summing the effects of changes in each variable: df = (∂f/∂x)dx + (∂f/∂y)dy
Higher-order derivatives involve taking derivatives of the partial derivatives. These derivatives help analyze complicated changes in multivariable functions. Consider a function f(x, y): - The first derivative is ∇f = (∂f/∂x, ∂f/∂y). - The second-order derivatives include ∂²f/∂x², ∂²f/∂y², and mixed partial derivatives like ∂²f/∂x∂y. - Higher-order derivatives continue this pattern, offering insight into the curvature and behavior of the function.
- At points where the function is not differentiable, partial derivatives may not exist. - For functions with discontinuities, derivatives are undefined at those points.
When dealing with derivatives of multivariable functions, students often encounter common pitfalls. Below are a few frequent mistakes and strategies to avoid them:
Calculate the gradient of f(x, y) = x²y + y²x at the point (1, 2).
The gradient of the function is ∇f = (∂f/∂x, ∂f/∂y). ∂f/∂x = 2xy + y² ∂f/∂y = x² + 2yx At the point (1, 2): ∂f/∂x = 2(1)(2) + 2² = 4 + 4 = 8 ∂f/∂y = 1² + 2(2)(1) = 1 + 4 = 5 Thus, the gradient at (1, 2) is (8, 5).
To find the gradient, calculate the partial derivatives with respect to x and y. Evaluate these derivatives at the given point to find the gradient vector.
A company uses a function f(x, y) = 3x² + 4y² to model cost based on two variables x and y. If x = 2 and y = 1, determine the rate of change of cost with respect to x.
The rate of change of the cost with respect to x is given by the partial derivative ∂f/∂x. ∂f/∂x = 6x At x = 2: ∂f/∂x = 6(2) = 12 Thus, the rate of change of cost with respect to x is 12.
Differentiate the cost function with respect to x to find the rate of change. Evaluate this derivative at the given value of x to determine the rate.
Find the second derivatives of the function f(x, y) = x³y².
First, find the first partial derivatives: ∂f/∂x = 3x²y² ∂f/∂y = 2x³y Second derivatives are: ∂²f/∂x² = 6xy² ∂²f/∂y² = 2x³ Mixed derivatives: ∂²f/∂x∂y = 6x²y ∂²f/∂y∂x = 6x²y Thus, the second derivatives are ∂²f/∂x² = 6xy², ∂²f/∂y² = 2x³, and ∂²f/∂x∂y = ∂²f/∂y∂x = 6x²y.
Calculate the first partial derivatives, then differentiate again to find the second derivatives, including mixed partial derivatives.
Prove that the mixed partial derivatives of f(x, y) = x²y + y²x are equal.
First, find the partial derivatives: ∂f/∂x = 2xy + y² ∂f/∂y = x² + 2yx Find the mixed partial derivatives: ∂²f/∂x∂y = ∂/∂y (2xy + y²) = 2x + 2y ∂²f/∂y∂x = ∂/∂x (x² + 2yx) = 2x + 2y Thus, ∂²f/∂x∂y = ∂²f/∂y∂x, proving they are equal.
Calculate the mixed partial derivatives by differentiating each partial derivative with respect to the other variable. Show that these derivatives are equal.
Evaluate the directional derivative of f(x, y) = x² + y² at the point (1, 1) in the direction of the vector (3, 4).
First, normalize the direction vector: u = (3, 4) |u| = √(3² + 4²) = 5 Normalized direction vector: u = (3/5, 4/5) Gradient of f: ∇f = (2x, 2y) At (1, 1): ∇f = (2, 2) Directional derivative = ∇f · u = (2, 2) · (3/5, 4/5) = (2)(3/5) + (2)(4/5) = 6/5 + 8/5 = 14/5 The directional derivative at (1, 1) is 14/5.
Normalize the direction vector, then compute the dot product of the gradient vector with the normalized direction vector to find the directional derivative.
- Partial Derivative: Derivative of a function with respect to one variable, holding others constant. - Gradient: A vector of partial derivatives indicating the direction of greatest increase of the function. - Directional Derivative: Rate of change of a function in a specific direction in its domain. - Mixed Partial Derivative: Derivative of a function with respect to two different variables. - Total Differentiation: Process of considering all partial derivatives to determine the overall rate of change of a function.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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