Chapter 10: Problem 16
Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=\left(x^{2}+y^{2}\right) e^{-x^{2}-y^{2}} ;(1,1)\)
Short Answer
Expert verified
Yes, \(f(x, y)\) is differentiable at \((1,1)\).
Step by step solution
01
Define Differentiability for Two Variables
A function \(f(x, y)\) is differentiable at a point \((a, b)\) if it has partial derivatives at \((a, b)\), and the linear approximation \( f(x, y) \approx f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \) holds when \((x, y)\) is near \((a, b)\). A function is differentiable at \((a, b)\) if the limit \[ \lim_{(h, k) \to (0, 0)} \frac{f(a + h, b + k) - f(a, b) - f_x(a, b)h - f_y(a, b)k}{\sqrt{h^2 + k^2}} = 0 \] exists.
02
Find Partial Derivatives
Calculate \(f_x(x, y)\) and \(f_y(x, y)\) using the product and chain rules. The partial derivative with respect to \(x\) is \[ f_x(x, y) = \frac{\partial}{\partial x} \left((x^2 + y^2) e^{-x^2 - y^2}\right) = 2xe^{-x^2-y^2} - 2x(x^2 + y^2)e^{-x^2 - y^2} \]Simplifying, we get, \[ f_x(x, y) = 2x e^{-x^2-y^2}(1 - x^2 - y^2) \]Similarly, for \(y\): \[ f_y(x, y) = \frac{\partial}{\partial y} \left((x^2 + y^2) e^{-x^2 - y^2}\right) = 2ye^{-x^2-y^2} - 2y(x^2 + y^2)e^{-x^2 - y^2} \]Simplifying, we get, \[ f_y(x, y) = 2y e^{-x^2-y^2}(1 - x^2 - y^2) \]
03
Evaluate the Partial Derivatives at (1, 1)
Substitute \((x, y) = (1, 1)\) into \(f_x(x, y)\) and \(f_y(x, y)\): \[ f_x(1, 1) = 2(1) e^{-1 - 1}(1 - 1 - 1) = 0 \]\[ f_y(1, 1) = 2(1) e^{-1 - 1}(1 - 1 - 1) = 0 \]
04
Use the Differentiability Criterion
As shown in Step 1, we need to check if the limit \[ \lim_{(h, k) \to (0, 0)} \frac{f(1 + h, 1 + k) - f(1, 1) - f_x(1, 1)h - f_y(1, 1)k}{\sqrt{h^2 + k^2}} = 0 \] exists.Since \(f_x(1, 1) = 0\) and \(f_y(1, 1) = 0\), this simplifies to \[ \lim_{(h, k) \to (0, 0)} \frac{f(1 + h, 1 + k) - f(1, 1)}{\sqrt{h^2 + k^2}} = 0 \]
05
Calculate \(f(1, 1)\) and \(f(1+h, 1+k)\)
Evaluate \(f(x, y)\) at \((1, 1)\): \[ f(1, 1) = (1^2 + 1^2) e^{-1^2 - 1^2} = 2e^{-2} \]For \((1+h, 1+k)\), use Taylor expansion: \[ f(1+h, 1+k) = ((1+h)^2 + (1+k)^2)e^{-(1+h)^2 - (1+k)^2} \approx (2 + 2h + 2k)e^{-2} + \text{higher-order terms} \] Higher-order terms vanish faster than \(\sqrt{h^2 + k^2}\).
06
Evaluate the Limit
Substitute the function values into the differentiability condition: \[ \lim_{(h, k) \to (0, 0)} \frac{(2 + 2h + 2k)e^{-2} - 2e^{-2}}{\sqrt{h^2 + k^2}} = \lim_{(h, k) \to (0, 0)} \frac{2he^{-2} + 2ke^{-2}}{\sqrt{h^2 + k^2}} = 0 \]Since the numerator becomes negligible when compared to \(\sqrt{h^2 + k^2}\), the limit is 0, proving differentiability.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Imagine you have a surface described by a function of two variables, like a smoothly sloping hillside. When we talk about partial derivatives, we're considering how the elevation of this surface changes as we move in specific directions, along the x-axis and the y-axis separately.
The partial derivative with respect to x, denoted as \(f_x(x, y)\), measures how the function changes as only the x variable is varied, keeping y constant. Similarly, \(f_y(x, y)\) measures changes of the function as only y varies.
The partial derivative with respect to x, denoted as \(f_x(x, y)\), measures how the function changes as only the x variable is varied, keeping y constant. Similarly, \(f_y(x, y)\) measures changes of the function as only y varies.
- **Product Rule:** Since the given function \(f(x, y) = (x^2 + y^2)e^{-x^2-y^2}\) is a product of two expressions, we use the product rule to find its derivative.
- **Chain Rule:** This rule helps when differentiating the exponential term \(e^{-x^2-y^2}\), since it involves another layer of functions.
Limits and Continuity
In calculus, particularly in multivariable calculus, limits help us understand the behavior of a function as it approaches a certain point. For a function to be continuous at a point, the function must be unbroken as it nears that point.
When demonstrating differentiability, we are actually checking for a specific type of limit condition. The criterion requires that the difference between the actual function value and a linear approximation becomes negligible as we get very close to the point.
When demonstrating differentiability, we are actually checking for a specific type of limit condition. The criterion requires that the difference between the actual function value and a linear approximation becomes negligible as we get very close to the point.
- This involves calculating a limit where both horizontal and vertical changes \((h, k)\) go to zero.
- The notation \(\lim_{(h, k) \to (0, 0)}\) denotes that we're considering the function's behavior in a multidirectional approach.
Multivariable Calculus
Multivariable calculus extends the principles of differentiation and integration to functions with more than one variable, such as \(f(x, y)\). Dealing with multiple variables allows for a richer analysis of how functions behave in a space rather than along a single line.
When determining differentiability, it's not only about finding derivatives but also understanding their significance in the context of a space formed by multiple dimensions.
When determining differentiability, it's not only about finding derivatives but also understanding their significance in the context of a space formed by multiple dimensions.
- **Visualization:** Imagine the function as a shape in three-dimensional space, where any movement results in a change both in direction and value.
- **Concept of Multidirectional Approach:** Here, we need to ensure the linear approximation works as it should in all directions around a point, making the calculation of limits crucial.