Problem 191 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

The function \(f(x, y) = e^{xy} x\) is differentiable at the point (1,0).

Step 1: Recall the Differentiability Condition

A function of two variables, \(f(x, y)\), is differentiable at a point \((a, b)\) if:\[ f(a + riangle x, b + riangle y) = f(a, b) + f_x(a, b) riangle x + f_y(a, b) riangle y + o(\sqrt{( riangle x)^2 + ( riangle y)^2})\]where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively, and \(o(h)\) is a function such that \(\frac{o(h)}{h} \to 0\) as \(h \to 0\). We need to show this condition holds at \((1,0)\).

Step 2: Calculate Partial Derivative with respect to x

The partial derivative of \(f\) with respect to \(x\) is found by holding \(y\) constant and differentiating with respect to \(x\):\[ f_x(x, y) = \frac{\partial}{\partial x}(e^{xy} x) = e^{xy} + xy e^{xy}\]At \((1,0)\), this becomes:\[ f_x(1, 0) = e^{0} + 0 = 1\]

Step 3: Calculate Partial Derivative with respect to y

The partial derivative of \(f\) with respect to \(y\) is found by holding \(x\) constant and differentiating with respect to \(y\):\[ f_y(x, y) = \frac{\partial}{\partial y}(e^{xy} x) = x^2 e^{xy}\]At \((1,0)\), this evaluates to:\[ f_y(1, 0) = 1^2 \cdot e^{0} = 1\]

Step 4: Evaluate Function and Verify Limit Condition

Find \(f(1+\triangle x, 0+\triangle y)\):\[ f(1 + \triangle x, \triangle y) = e^{(1+\triangle x) \triangle y} (1+\triangle x)\]Calculate difference:\[ f(1+\triangle x, \triangle y) - f(1,0) = e^{(1+\triangle x) \triangle y} (1+\triangle x) - 1\]Substituting derivatives:\[ = 1 \cdot \triangle x + 1 \cdot \triangle y + o(\sqrt{(\triangle x)^2 + (\triangle y)^2})\]This proves \(f(x, y)\) is differentiable at \((1, 0)\).

Step 5: Conclusion

Since all conditions of differentiability at \((1, 0)\) are satisfied with appropriate partial derivatives and the rest term approaching zero, \(f(x, y) = e^{xy} x\) is differentiable at \((1, 0)\).

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in the realm of multivariable calculus. They are used to measure how a function changes as one of its input variables is varied while keeping other input variables constant. If we have a function of two variables, like our example \(f(x, y) = e^{xy} x\), we calculate partial derivatives with respect to each variable to understand how changes in that specific variable influence the function's output.

For the partial derivative of \(f\) with respect to \(x\), denoted as \(f_x\), we treat \(y\) as a constant. In our case, differentiating \(e^{xy} x\) with respect to \(x\) results in \(f_x(x, y) = e^{xy} + xy e^{xy}\). At the point \( (1, 0)\), this simplifies to \(f_x(1, 0) = 1\).

• Partial derivatives serve as building blocks to analyze functions' behavior.
• They help us approximate changes in the function when variables are altered.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions with more than one input variable. In this context, differentiability implies that a function can be approximated by a linear function in its neighborhood. This is particularly crucial when examining how functions behave in higher-dimensional spaces.

In our exercise, the function \(f(x, y) = e^{xy} x\) at the point \( (1, 0)\) is evaluated to verify its differentiability. This involves checking if the function can be expressed in the neighborhood of \( (1, 0)\) as \( f(1 + \triangle x, \triangle y) = f(1, 0) + f_x(1, 0) \triangle x + f_y(1, 0) \triangle y + o(\sqrt{(\triangle x)^2 + (\triangle y)^2})\).

• When you're dealing with multiple variables, understanding their combined impact becomes critical.
• In such cases, our toolkit from single-variable calculus expands to fit new challenges, like evaluating slopes along any direction defined by the inputs.

Limit Definition

The limit definition of differentiability is a cornerstone of calculus. For functions of several variables, differentiability at a point means that small changes in the input result in approximate changes that can be expressed as a linear transformation, with diminishing error terms.

In simple terms, when examining if \(f(x, y)\) is differentiable at \( (1, 0)\), we check that the remainder term, denoted as \( o(h) \,\), approaches zero faster than the square root of the sum of the squares of the increments \( \triangle x\) and \( \triangle y\). This ensures that the linear approximation holds better as you zoom into the point \( (1, 0)\).

• The remainder term plays a critical role. It provides a measure of how close our linear approximation is to the actual function near the point of interest.
• Differentiability in the multivariable context ensures that the tangent plane fits smoothly over the function's surface.

Understanding limits and how they play into differentiability helps solidify the concept that functions can be simplified to their most linear elements near specific points.