Question

If \(z=x y+x+y, x=r+s+t\), and \(y=r s t\), find \(\left.\frac{\partial z}{\partial s}\right|_{r=1, s=-1, t=2}\)

Short answer

The partial derivative \( \frac{\partial z}{\partial s} \) evaluated at \( r=1, s=-1, t=2 \) is 1.

Step-by-step solution

Understand the Partial Derivative

We're asked to find the partial derivative of \( z \) with respect to \( s \), denoted as \( \frac{\partial z}{\partial s} \). This involves treating all other variables (\( r \), \( t \), and any functions of them) as constants.

Substitute Expressions for x and y

Since \( x = r + s + t \) and \( y = r s t \), substitute these into the expression for \( z \): \( z = (r + s + t)(r s t) + (r + s + t) + r s t \).

Expand the Expression for z

Expand the expression for \( z \). First, multiply: \((r+s+t)(r s t) = r^2st + rs^2t + rst^2\). Next, combine to get: \( z = r^2st + rs^2t + rst^2 + r + s + t + rst \).

Find the Partial Derivative of z with Respect to s

Take the derivative of \( z \) with respect to \( s \). Using power and constant rules, derive: \( \frac{\partial z}{\partial s} = 2rst + rt^2 + 1 \).

Evaluate the Partial Derivative at Given Values

Substitute \( r = 1 \), \( s = -1 \), and \( t = 2 \) into \( \frac{\partial z}{\partial s} = 2rst + rt^2 + 1 \) giving \( 2(1)(-1)(2) + (1)(4) + 1 = -4 + 4 + 1 = 1 \).

Conclusion

The value of the partial derivative \( \frac{\partial z}{\partial s} \) evaluated at \( r=1 \), \( s=-1 \), and \( t=2 \) is \( 1 \).

Key concepts

Multivariable Calculus

Multivariable calculus is essential when dealing with functions that depend on more than one variable. In this context, we encounter functions that are expressed in terms of several different variables, such as in the given exercise where we have functions of three variables: \(x\), \(y\), and \(z\).
These variables themselves depend on other variables \(r\), \(s\), and \(t\). The central goal of multivariable calculus is to explore how changes in these variables affect the function.

• Partial derivatives: We use partial derivatives to determine how a function changes with respect to one of its multiple variables, holding the other variables constant.
• Simplifying complex systems: Techniques in multivariable calculus help simplify and solve real-world problems involving multiple changing quantities.

Multivariable calculus bridges simple calculus with more applied mathematical concepts, paving the way for complex engineering and scientific applications.

Chain Rule

The chain rule in calculus is a formula used to compute the derivative of a composite function. When you have a function that depends on a set of variables, each of which might itself be a function of another variable, you would use the chain rule to find the derivative.
In our exercise, the function \(z\) depends on \(x\) and \(y\), which in turn are functions of \(r\), \(s\), and \(t\).
This is how it plays out:

• Simple dependencies: The chain rule helps take derivatives in layers, following paths of dependency through each function layer. This is akin to peeling an onion, layer by layer.
• Critical in differentiation: Particularly important when dealing with nested functions in multivariable calculus problems, where variables are interdependent.

Applying the chain rule carefully allows us to handle complex derivations and find solutions in intricate systems of functions.

Function Expansion

Function expansion involves rewriting a function in terms of its base components, often to simplify differentiation. In the original problem, we expand \(z = (r + s + t)(r s t) + (r + s + t) + r s t\), simplifying the task of taking derivatives.
This approach includes:

• Breaking down expressions: Express the interactions, such as multiplying terms, into simpler, visible components. For example, we expanded \((r + s + t)(r s t)\) into individual sums of products.
• Facilitating calculations: Expansion simplifies complex mathematical operations by making each term explicit.

By expanding functions, differentiation and other derivative computations become manageable, preventing errors and oversights in calculations.

Differentiation

Differentiation allows us to find the rate at which a function changes as its input variables change. In the context of multivariable calculus, it's critical to isolate how specific variables affect a function among many others.
For instance, in our problem, we compute the derivative of \(z\) with respect to \(s\). This is known as a partial derivative, written as \( \frac{\partial z}{\partial s} \).

• Partial derivatives: Indicate the function's sensitivity to changes in a particular direction or along a specific variable axis.
• Utilizing rules: Applying standard differentiation rules to each term in the expanded function results, as seen with power and constant rules.

These techniques not only measure propensity for change but also help us explore connections between variables in a system.

Substitution in Functions

Substitution in calculus involves replacing variables with expressions or numerical values to simplify the problem-solving process. In this exercise, substitutions were used at various stages:
The substitutions \(x = r + s + t\) and \(y = r s t\) transform the function \(z\) into a form suitable for differentiation.
Here’s why substitution is useful:

• Simplifying expressions: By substituting, we manage and reduce complex interdependencies among variables.
• Navigating composite systems: Provides clarity and cohesion in functions that initially appear convoluted.

Finally, values \(r=1\), \(s=-1\), and \(t=2\) were substituted into the derivative to evaluate it. This step simplifies calculating specific solutions, making substitution an essential part of working with multivariable functions.