Problem 23 - Calculus with Concepts in Calculus

Question

Find \(\partial w / \partial u\) and \(\partial w / \partial v\). $$ w=y \ln x z ; x=v e^{u}, y=u^{2} v^{4}, z=u e^{v} $$

Step-by-Step Solution

Answer

\(\partial w / \partial u = v^4 (2u \ln(vu) + 2u^2 + 2uv + u)\), \(\partial w / \partial v = u^2 (4v^3 \ln(vu) + v^3(1 + u + v))\).

Step 1: Substitute the Expressions

Begin by substituting the given expressions for \(x\), \(y\), and \(z\) into the equation for \(w\). This yields: \[ w = (u^2 v^4) \ln((v e^u) (u e^v)). \]

Step 2: Simplify the Logarithm

Simplify the equation for \(w\) by multiplying inside the logarithm: \[ w = u^2 v^4 \ln ((vu)(e^{u+v})). \] This simplifies to: \[ w = u^2 v^4 (\ln(vu) + (u+v)). \]

Step 3: Find Partial Derivative \(\partial w / \partial u\)

To find \(\partial w / \partial u\), apply the product and chain rule to differentiate \( w \) with respect to \( u \): \[ \frac{\partial w}{\partial u} = 2u v^4 (\ln(vu) + (u+v)) + u^2 v^4 \left(\frac{1}{u} + 1\right). \] Simplifying yields: \[ \frac{\partial w}{\partial u} = v^4 (2u \ln(vu) + 2u^2 + 2uv + u). \]

Step 4: Find Partial Derivative \(\partial w / \partial v\)

Now, find \(\partial w / \partial v\) by differentiating \( w \) with respect to \( v \): \[ \frac{\partial w}{\partial v} = u^2 (4v^3 \ln(vu) + v^4 \left(\frac{1}{v} + u + 1\right)). \] Simplifying gives: \[ \frac{\partial w}{\partial v} = u^2 (4v^3 \ln(vu) + v^3(1 + u + v)). \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when working with functions of several variables. It helps us to differentiate composite functions. In a nutshell, when you have a function inside another function, you use the chain rule to find the derivative.

For example, consider a composite function like \( f(g(x)) \). The chain rule says that the derivative \( f'(g(x)) \) is the product of the derivative of \( f \) with respect to \( g \) and the derivative of \( g \) with respect to \( x \). Mathematically, this can be written as:

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

In the context of the given exercise, where \( w = y \ln(xz) \) and \( x, y, \) and \( z \) are functions of \( u \) and \( v \), the chain rule helps us differentiate with respect to these variables.

When differentiating \( w \) with respect to \( u \), the chain rule assists in accounting for how changes in \( u \) affect \( y \), \( \ln(xz) \), and other nested functions, ensuring we consider each part's contribution to the rate of change of \( w \). "Inside" functions like \( x = ve^u \) and \( z = ue^v \) change based on \( u \), and the chain rule tells us how their derivatives feed into the larger picture.

Product Rule

The product rule is another cornerstone in calculus when dealing with derivatives. It is essential when the function you need to differentiate is a product of two or more other functions. The rule tells us how to derive the product of two functions correctly.

For a product of two functions \( u(x) \) and \( v(x) \), the product rule is applied as follows:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

This principle applies directly to our exercise. Consider \( w = u^2 v^4 (\ln(vu) + (u+v)) \). When finding the partial derivative \( \partial w / \partial u \), the product rule helps us differentiate each piece accurately.

By splitting the differentiation across the components of \( w \), where one part could affect another's derivative, the product rule ensures that every term is accounted for and differentiated accordingly. This results in an accurate calculation of how partial derivatives of \( w \) with respect to \( u \) are determined.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of more than one variable. In contrast to simpler functions of a single variable, multivariable functions can vary based on several inputs. Thus, derivatives become partial, representing how a change in one variable affects the overall function, while keeping others constant.

The given exercise showcases this by dealing with a function \( w \) that is not only dependent on \( u \) and \( v \) but also involves intermediate functions like \( x, y, \) and \( z \). Here, each of these expressions contributes dynamically based on changes to \( u \) or \( v \).

• Partial derivatives reflect how \( w \) changes with each independent variable.
• Chain rule and product rule are tools that cater to these multivariable functions.

In practice, knowing how to find these partial derivatives is useful in fields like physics and engineering, where variables frequently interact in complex ways. In this exercise, we distinguished between partial derivatives \( \partial w / \partial u \) and \( \partial w / \partial v \), taking into account specific dependencies on each variable. This approach allows a deeper understanding of how each variable separately influences the function.