Problem 25 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$f(r, s)=r \ln \left(r^{2}+s^{2}\right)$$

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial r} = \ln(r^2 + s^2) + \frac{2r^2}{r^2 + s^2}\), \(\frac{\partial f}{\partial s} = \frac{2rs}{r^2 + s^2}\)

Step 1: Understanding the Function

We have the function \( f(r, s) = r \ln(r^2 + s^2) \). We need to find the first partial derivatives with respect to \( r \) and \( s \). This means treating one variable as a constant while differentiating with respect to the other.

Step 2: Finding the Partial Derivative with respect to \( r \)

To find \( \frac{\partial f}{\partial r} \), treat \( s \) as a constant. Using the product rule, which states that \((uv)' = u'v + uv'\), we have: \( f(r, s) = u \cdot v \) where \( u = r \) and \( v = \ln(r^2 + s^2) \). First, find \( u' = \frac{d}{dr}(r) = 1 \) and \( v' = \frac{d}{dr}(\ln(r^2 + s^2)) = \frac{2r}{r^2 + s^2} \). So, \( \frac{\partial f}{\partial r} = (1)\ln(r^2+s^2) + r\left(\frac{2r}{r^2 + s^2}\right) = \ln(r^2 + s^2) + \frac{2r^2}{r^2 + s^2} \).

Step 3: Finding the Partial Derivative with respect to \( s \)

To find \( \frac{\partial f}{\partial s} \), treat \( r \) as a constant. The function becomes \( f(s) = r \ln(r^2 + s^2) \). Applying the chain rule, \( \frac{d}{ds}(r \ln(r^2 + s^2)) = r \cdot \frac{d}{ds}(\ln(r^2 + s^2)) = r \cdot \frac{2s}{r^2 + s^2} \). Therefore, \( \frac{\partial f}{\partial s} = \frac{2rs}{r^2 + s^2} \).

Key Concepts

Understanding the Product Rule

In calculus, the product rule is essential when dealing with the differentiation of functions that are products of two or more sub-functions. This rule states that if you have two functions, say \( u(x) \) and \( v(x) \), their derivative is given by:

• \( (uv)' = u'v + uv' \)

This means you differentiate the first function and multiply it by the second, and then do the reverse: keep the first function and differentiate the second. In our original exercise in step 2, we identified our function as a product where \( u = r \) and \( v = \ln(r^2 + s^2) \). We then applied the product rule to find the partial derivative with respect to \( r \). By treating "\( v \)" as a function dependent on \( r \), we simplified and found our derivative.
The product rule is crucial when functions are intertwined in a multiplication, which is common in multivariable calculus.

The Versatility of the Chain Rule

The chain rule is another foundational concept in calculus, especially when dealing with composite functions. It allows us to differentiate functions nested within each other. A simple way to think of it is: if you have a "function within a function," or a composite function \( y = f(g(x)) \), the derivative is:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In our exercise, while calculating \( \frac{\partial f}{\partial s} \), the chain rule was used on the function \( \ln(r^2 + s^2) \). Here, "\( r^2 + s^2 \)" is nested within the natural logarithm function. We find the derivative of the outer function, by considering the derivative of the inner part, and multiply these results. Thus, by treating \( r \) as a constant and focusing on the \( s \)-component, we applied the chain rule to arrive at the partial derivative.
The chain rule simplifies complex differentiation tasks and is indispensable in multivariable calculus for breaking down intricate problems.

An Introduction to Multivariable Calculus

Multivariable calculus extends the realm of single-variable calculus into functions with more than one variable. This allows us to explore systems where quantities are influenced by multiple factors. Unlike single-variable calculus, the function's output depends on several inputs. Therefore, partial derivatives come into play, highlighting how the function changes concerning one variable while keeping others constant.
In the exercise you've seen, the function \( f(r, s) = r \ln(r^2 + s^2) \) depends on both \( r \) and \( s \).

• By using partial derivatives, we successfully evaluated how \( f \) changes with "only \( r \)" and "only \( s \)."

This approach not only helps in understanding how a particular variable influences the function but is also practical in real-life scenarios where outcomes depend on multiple variables like in physics, economics, and engineering. Multivariable calculus remains a cornerstone of higher mathematics with widespread applications.