Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(F(x, y)=\ln \left(x^{2}+x y+y^{2}\right)\), find \(F_{x}(-1,4)\) and \(F_{y}(-1,4)\)

Step-by-Step Solution

Answer

\( F_x(-1,4) = \frac{2}{13} \) and \( F_y(-1,4) = \frac{7}{13} \).

Step 1: Understand the Problem

We need to find the partial derivatives \( F_x(x, y) \) and \( F_y(x, y) \) of the function \( F(x, y)=\ln (x^2+xy+y^2) \), and then evaluate these derivatives at the point \((-1, 4)\).

Step 2: Find Partial Derivative with Respect to x

To find \( F_x(x, y) \), use the chain rule. Start by setting \( u = x^2 + xy + y^2 \). Then \( F(x, y) = \ln u \), so \( F_x = \frac{1}{u} \cdot \frac{\partial u}{\partial x} \). We have \( u = x^2 + xy + y^2 \), so \( \frac{\partial u}{\partial x} = 2x + y \). Therefore, \( F_x(x, y) = \frac{2x + y}{x^2 + xy + y^2} \).

Step 3: Evaluate \( F_x(-1, 4) \)

Substitute \( x = -1 \) and \( y = 4 \) into the expression for \( F_x(x, y) \). Find \( F_x(-1, 4) = \frac{2(-1) + 4}{(-1)^2 + (-1)(4) + 4^2} = \frac{-2 + 4}{1 - 4 + 16} = \frac{2}{13} \).

Step 4: Find Partial Derivative with Respect to y

To find \( F_y(x, y) \), again use the chain rule. Start with \( u = x^2 + xy + y^2 \) so \( F(x, y) = \ln u \), therefore \( F_y = \frac{1}{u} \cdot \frac{\partial u}{\partial y} \). Here, \( \frac{\partial u}{\partial y} = x + 2y \). Thus, \( F_y(x, y) = \frac{x + 2y}{x^2 + xy + y^2} \).

Step 5: Evaluate \( F_y(-1, 4) \)

Substitute \( x = -1 \) and \( y = 4 \) into the expression for \( F_y(x, y) \). Find \( F_y(-1, 4) = \frac{-1 + 2(4)}{(-1)^2 + (-1)(4) + 4^2} = \frac{-1 + 8}{1 - 4 + 16} = \frac{7}{13} \).

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation rule used frequently in calculus, especially when dealing with composite functions. It's crucial when working with functions depending on multiple variables, as seen in partial derivatives.

In the context of the exercise, the function is given by \[F(x, y) = \ln(x^2 + xy + y^2)\] and to differentiate this with respect to one variable, we use the chain rule. Here's how it helps:

• Firstly, identify the inner function \(u = x^2 + xy + y^2\), which means \(F(x, y) = \ln u\).
• To find the partial derivative \(F_x\) or \(F_y\), differentiate the outer function (\(\ln(u)\)) with respect to \(u\), giving \(\frac{1}{u}\).
• Next, find the derivative of the inner function with respect to the desired variable: \(\frac{\partial u}{\partial x} = 2x + y\) for \(x\) and \(\frac{\partial u}{\partial y} = x + 2y\) for \(y\).

Using the chain rule, you can then reach the partial derivatives \(F_x\) and \(F_y\), expressed in terms of \(u\). This method simplifies evaluating derivatives for complex multivariable functions.

Evaluation at a Point

Once you've found the partial derivatives, evaluating them at a specific point involves substituting the coordinates into the derivative expressions.

For the problem, we need to evaluate both \(F_x(x, y)\) and \(F_y(x, y)\) at the point \((-1, 4)\). Here's the approach broken down:

• Start with the partial derivatives found earlier: \(F_x(x, y) = \frac{2x + y}{x^2 + xy + y^2}\) and \(F_y(x, y) = \frac{x + 2y}{x^2 + xy + y^2}\).
• Substitute \(x = -1\) and \(y = 4\) into these expressions.
• For \(F_x(-1, 4)\), you calculate \(\frac{-2 + 4}{1 - 4 + 16} = \frac{2}{13}\).
• Similarly, for \(F_y(-1, 4)\), you compute \(\frac{-1 + 8}{1 - 4 + 16} = \frac{7}{13}\).

Evaluating at a point gives specific insights into the behavior and slope of the function's graph at that particular location.

Logarithmic Functions

Logarithmic functions play a vital role in many areas of mathematics, particularly in calculus for differentiating composite expressions like in our exercise function \(F(x, y) = \ln(x^2 + xy + y^2)\).

Understanding their properties is key:

• The derivative of \(\ln(u)\) with respect to \(u\) is simply \(\frac{1}{u}\). This fundamental rule makes logarithmic differentiation straightforward.
• Logarithms are useful because they transform multiplicative relationships into additive ones, simplifying complex expressions before differentiation.
• In the exercise, since \(F(x, y)\) involves \(\ln(u)\), calculating its partial derivatives is efficient using the chain rule.

Knowing how to handle logarithmic functions allows for precise calculation of derivatives, especially when dealing with products, quotients, or powers embedded within logarithmic expressions.