Problem 46 - Advanced Engineering Mathematics

Question

Assume that \(f\) and \(g\) are differentiable functions of two variables. Prove the given identity. $$ \nabla(f+g)=\nabla f+\nabla g $$

Step-by-Step Solution

Answer

The gradient of the sum of functions equals the sum of their gradients.

Step 1: Understand the Problem

The problem asks us to prove the identity \( abla (f+g) = abla f + abla g \), where \( f \) and \( g \) are differentiable functions of two variables. This means we need to show that the gradient of the sum of two functions equals the sum of their individual gradients.

Step 2: Define the Gradient

The gradient of a scalar function \( f(x, y) \) is a vector, given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Similarly, for \( g(x, y) \), \( abla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) \).

Step 3: Apply the Gradient to the Sum

Find the gradient of the function \( f+g \). Using the definition, \( abla (f+g) = \left( \frac{\partial (f+g)}{\partial x}, \frac{\partial (f+g)}{\partial y} \right) \).

Step 4: Use Linearity of Partial Derivatives

Since partial derivatives are linear operators, \( \frac{\partial (f+g)}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x} \) and \( \frac{\partial (f+g)}{\partial y} = \frac{\partial f}{\partial y} + \frac{\partial g}{\partial y} \).

Step 5: Combine Results to Form the Proof

Substitute these results into the gradient of the sum: \( abla (f+g) = \left( \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x}, \frac{\partial f}{\partial y} + \frac{\partial g}{\partial y} \right) \). This equals \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) + \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = abla f + abla g \).

Step 6: Conclude the Proof

We have shown that \( abla (f+g) = abla f + abla g \), thus proving the identity.

Key Concepts

Differentiable Functions

In calculus, differentiable functions play a crucial role. A function is called differentiable at a point if it has a derivative at that point, and it is differentiable over a region if it has derivatives at every point in that region. For a function of two variables, being differentiable essentially means that the function behaves well around every point it is defined on. Differentiability implies that the function can be locally approximated by a linear function, which is great for analyzing the behavior of complex functions. For example, if you have functions like \( f(x, y) \) and \( g(x, y) \), understanding their differentiability allows you to know that they have well-defined derivatives with respect to both \( x \) and \( y \). This concept is foundational, as it assures us that we can use certain mathematical tools to examine these functions, such as gradients and partial derivatives. This is exactly what happens when you try to find the gradient of the sum \((f+g)\) when \( f \) and \( g \) are differentiable functions.

Partial Derivatives

Partial derivatives are a tool used in calculus to understand how functions change. They focus on one variable at a time, keeping others constant, which simplifies the understanding of functions with more than one variable. For a function \( f(x, y) \), the partial derivative with respect to \( x \) is noted as \( \frac{\partial f}{\partial x} \), and it tells us how \( f \) changes as \( x \) changes while \( y \) is held constant. The same goes for partial derivative with respect to \( y \), noted as \( \frac{\partial f}{\partial y} \). By finding the partial derivatives of both functions \( f \) and \( g \) in relation to both variables, we can effectively understand the behavior of these functions when summed together, as in \( (f+g) \). That's precisely what the problem is about: showing that these partial derivatives are additive due to the linearity of differential operations.

Linearity of Operators

The linearity of operators is an important property in mathematics, especially when dealing with derivatives and integrals. Simply put, an operator is linear if it satisfies two conditions: additivity and homogeneity. For derivatives, linearity means that the derivative of a sum is the sum of the derivatives, and the derivative of a constant times a function is the constant times the derivative of the function. In the context of partial derivatives, this means:

• \( \frac{\partial (f+g)}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x} \)
• \( \frac{\partial (f+g)}{\partial y} = \frac{\partial f}{\partial y} + \frac{\partial g}{\partial y} \)

This property is what we rely on to prove that the gradient of a sum, \( abla (f+g) \), is equal to the sum of the gradients, \( abla f + abla g \). It's the backbone of our problem's solution. By using this inherent linearity, complex expressions can be broken down and simplified, making them easier to work with, especially in multi-variable calculus.