Problem 191 - Calculus 3

Question

Show that \(f(x, y)=e^{x y} x\) is differentiable at point \((1,0) .\)

Step-by-Step Solution

Answer

The function is differentiable at point (1,0).

Step 1: Understand the Differentiability Criterion

A function of two variables, \(f(x,y)\), is differentiable at a point \((a,b)\) if it can be approximated by a linear function near that point. This means \(f(a + h, b + k) \approx f(a, b) + f_x(a, b)h + f_y(a, b)k\) where \(h\) and \(k\) are small and \(f_x\) and \(f_y\) are the partial derivatives of \(f\) at \((a,b)\).

Step 2: Calculate Partial Derivative with Respect to x

First, find the partial derivative of \(f(x, y) = e^{xy} x\) with respect to \(x\). Use the product and chain rules: \( f_x(x, y) = \frac{\partial}{\partial x}(e^{xy} x) = e^{xy} + e^{xy} x y\). Evaluate at \((1, 0)\) to get \(f_x(1, 0) = e^0 = 1\).

Step 3: Calculate Partial Derivative with Respect to y

Next, find the partial derivative with respect to \(y\). This is \( f_y(x, y) = \frac{\partial}{\partial y}(e^{xy} x) = e^{xy} x^2\). Evaluate at \((1, 0)\) to get \(f_y(1, 0) = e^0 \cdot 1^2 = 1\).

Step 4: Check Limit of Linear Approximation Error

Define the error function \(E(h,k)\) by \(E(h,k) = \frac{f(1+h, k) - f(1, 0) - f_x(1,0)h - f_y(1,0)k}{\sqrt{h^2 + k^2}}\). Evaluate \(f(1+h, k) = e^{(1+h)k}(1+h) = e^k + he^k + k + hk\). Hence, the error term is \(E(h,k) = \frac{he^k + hk}{\sqrt{h^2 + k^2}}\). As \((h,k)\) approaches \((0,0)\), \(E(h,k)\) approaches zero.

Step 5: Conclusion on Differentiability

Since the error \(E(h,k)\) approaches zero as \((h,k)\) approaches \( (0,0)\), the function \(f(x, y)=e^{x y} x\) is differentiable at the point \((1,0)\).

Key Concepts

Partial Derivative

To understand the concept of partial derivatives, imagine a function that takes multiple inputs, like our function \( f(x, y) = e^{xy} x \). Unlike a function of a single variable, this one involves two variables affecting the outcome. A partial derivative focuses on how a function changes as we vary one input at a time while keeping the others constant.
For example, the partial derivative of \( f \) with respect to \( x \) is noted as \( f_x \). This involves treating \( y \) as a constant when differentiating with respect to \( x \). Similarly, \( f_y \) measures the change in \( f \) as \( y \) varies while \( x \) stays fixed.

• To find these, use derivative rules like the product rule: \( (uv)' = u'v + uv' \)
• Apply the chain rule when dealing with compositions of functions.

For our function, we calculated \( f_x(x, y) = e^{xy} + e^{xy}xy \) and \( f_y(x, y) = e^{xy} x^2 \). It is handy to evaluate at specific points, like \((1, 0)\), to understand the function's behavior at that location.
Partial derivatives are instrumental for assessing how a function behaves at any given point, laying the groundwork for concepts like the differentiability of multivariable functions.

Linear Approximation

Linear approximation is a powerful technique that allows us to approximate the behavior of functions using a simple linear equation. Think of it as zooming in on a curve until it looks almost like a straight line. For functions of two variables, this is akin to finding a plane that closely fits the surface near a given point.
The formula for linear approximation near a point \((a, b)\) involves the function value at that point and its partial derivatives: \[ f(a + h, b + k) \approx f(a, b) + f_x(a, b)h + f_y(a, b)k \]This approximation works best when \(h\) and \(k\) are small, capturing the linear behavior around the point.

• It's valuable for simplifying complex functions into something more easily analyzed.
• Linear approximations are frequently the first step in iterative numerical methods.

In our example, we used this linear form to verify differentiability. By examining the error term and ensuring it approaches zero as \((h, k)\) shrink to zero, we confirm that \( f(x, y) = e^{xy} x \) indeed mimics linear behavior at \((1, 0)\), proving it is differentiable there.

Chain Rule

The chain rule is crucial when dealing with the differentiation of composite functions. It answers how we can differentiate a function that depends on another function. In practical terms, it helps deconstruct these layers of functions into manageable parts.
For functions like \(f(x, y) = e^{xy} x\), we often encounter terms where one variable affects another through a composite relationship. The chain rule enables differentiation by considering each component's contribution and then combining them.

• If you have a function \( u(g(x)) \), the derivative with respect to \( x \) is \( du/dg \cdot dg/dx \).
• This rule is applied repeatedly in multivariable calculus, especially when functions are not purely independent of one another.

In our provided solution, the chain rule was used together with the product rule to differentiate terms like \( e^{xy} \) with respect to both \( x \) and \( y \). Understanding and applying these rules correctly is essential in calculus, especially for verifying differentiability of functions with multiple variables. It simplifies differentiation into digestible parts, unveiling how changes propagate in interconnected systems.