Problem 28 - Calculus an Applied Approach

Question

Evaluate \(f_{x}\) and \(f_{y}\) at the point. $$ \text { Function } \quad \text { Point } $$ $$ f(x, y)=\ln \sqrt{x y} \quad(-1,-1) $$

Step-by-Step Solution

Answer

The partial derivatives \(f_x\) and \(f_y\) of the function at the point (-1,-1) are both equal to -1/2.

Step 1: Compute \(f_x\)

The derivative of \(f\) with respect to \(x\) is obtained by differentiating the function while considering \(y\) as constant. Here, the chain rule is used because the function is a composition of the natural log function and the square root function, both of which depend on \(x\). The chain rule states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function.\[f_x = \frac{1}{\sqrt{xy}}*\frac{1}{2\sqrt{x}}=\frac{1}{2x\sqrt{y}}\]

Step 2: Compute \(f_y\)

The derivative of \(f\) with respect to \(y\) is obtained by differentiating the function while considering \(x\) as constant. Again, the chain rule is used due to the composition of the function.\[f_y = \frac{1}{\sqrt{xy}}*\frac{1}{2\sqrt{y}}=\frac{1}{2y\sqrt{x}}\]

Step 3: Evaluate at the point (-1,-1)

Substitute the point into the partial derivatives. For \(f_x\), substitute (-1,-1) into \(\frac{1}{2x\sqrt{y}}\) to get \(\frac{1}{2*(-1)*\sqrt{-1}}= -\frac{1}{2}\). For \(f_y\), substitute (-1,-1) into \(\frac{1}{2y\sqrt{x}}\) to get \(\frac{1}{2*(-1)*\sqrt{-1}}= -\frac{1}{2}\).

Key Concepts

Chain Rule

The chain rule is a powerful method in calculus for finding the derivative of composite functions. A composite function is a function that is formed by combining two or more functions, such that one function is applied to the result of another. For example, in the function \(f(x, y) = \ln \sqrt{xy}\), there is a composition of the natural logarithm and the square root functions.
The chain rule allows us to differentiate these complex functions by breaking them down into their constituent parts. The rule states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function.

• Outer Function: In our example, it is the natural logarithm \(\ln(z)\), where \(z\) is the inner function \(\sqrt{xy}\).
• Inner Function: The inner function is \(\sqrt{xy}\), which itself involves the multiplication of \(x\) and \(y\), followed by a square root.

To compute \(f_x\), you treat \(y\) as a constant and differentiate with respect to \(x\), using the chain rule. Similarly, to find \(f_y\), treat \(x\) as a constant and differentiate with respect to \(y\). By following this method, you can handle quite complex derivatives efficiently.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. The natural logarithm function has many important properties and applications, especially in calculus and mathematical analysis.
The derivative of the natural logarithm \(\ln(x)\) is particularly useful and is given by \(\frac{1}{x}\) when \(x > 0\). However, when dealing with composites involving the logarithm, you apply the chain rule. For instance, the function \(\ln\sqrt{xy}\) requires additional steps to handle the embedded functions.
In the context of the exercise, differentiating the function involves taking the derivative of \(\ln(u)\) where \(u = \sqrt{xy}\). Using the chain rule:

• Step 1: Identify the outer function \(\ln(u)\), which differentiates to \(\frac{1}{u}\).
• Step 2: Determine and differentiate the inner function \(u = \sqrt{xy}\).

By following these steps, you can handle functions involving natural logs systematically and accurately.

Square Root Function

The square root function often appears in calculus problems and is symbolized by \(\sqrt{x}\). It maps a non-negative number \(x\) to another non-negative number \(y\) such that \(y^2 = x\). It's an essential component of many complex functions.
In differentiation, the square root function \(\sqrt{x}\) can be rewritten as \(x^{1/2}\). This exponent form is often easier to work with, especially when using rules of differentiation.

• The derivative of \(\sqrt{x}\), or \(x^{1/2}\), is \(\frac{1}{2}x^{-1/2}\).
• When the square root is within another function, you typically need to use the chain rule.

For the given function \(f(x, y) = \ln \sqrt{xy}\), the square root \(\sqrt{xy}\) is part of the inner function. To differentiate with respect to \(x\), for example, the expression \(\sqrt{xy}\) is treated as \((xy)^{1/2}\), and you apply the power rule as part of the chain rule.
In this way, understanding the square root function and its role in differentiation empowers you to tackle a wide range of calculus problems.