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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

Expert verified
The answer is 1.

Step by step solution

01

Understand the Definitions

We start with the problem definitions: \(z = xy + x + y\), \(x = r + s + t\), and \(y = rst\). Our goal is to find \( \left. \frac{\partial z}{\partial s} \right|_{r=1, s=-1, t=2} \). This means we need to compute the partial derivative of \(z\) with respect to \(s\), and then evaluate it at \(r = 1\), \(s = -1\), and \(t = 2\).
02

Substitute Definitions Into Original Function

Substitute \(x = r + s + t\) and \(y = rst\) into the function \(z = xy + x + y\). This gives us: \[z = [(r+s+t)(rst)] + (r+s+t) + (rst)\]
03

Simplify the Expression for z

Simplify the expression:1. First, expand \((r+s+t)(rst) = r^2st + srst + trst\).2. Combine and simplify terms:\[z = r^2st + srst + trst + r + s + t + rst\]Which simplifies to:\[z = r^2st + srst + trst + rst + r + s + t\]
04

Differentiate z With Respect to s

To find \(\frac{\partial z}{\partial s}\), take the partial derivative of \(z\) with respect to \(s\):\[\frac{\partial z}{\partial s} = \frac{\partial }{\partial s}(r^2st + srst + trst + rst + r + s + t)\]Each term is derived as follows:- \(r^2st\) with respect to \(s\) is \(r^2t\).- \(srst\) with respect to \(s\) is \(rst\).- \(trst\) is treated as a constant with respect to \(s\).- \(rst\) is a constant with respect to \(s\).- \(r\), \(t\) are constants with respect to \(s\).- \(s\) with respect to \(s\) is 1.Thus,\[\frac{\partial z}{\partial s} = r^2t + rst + 1\]
05

Evaluate the Partial Derivative at Given Values

Evaluate \(\frac{\partial z}{\partial s}\) at \(r = 1\), \(s = -1\), and \(t = 2\): \[\frac{\partial z}{\partial s}\mid_{r=1,s=-1,t=2} = (1)^2(2) + (1)(-1)(2) + 1 \]Simplify:\[= 2 - 2 + 1 = 1\]
06

Conclusion

The final result of the partial derivative \(\left.\frac{\partial z}{\partial s}\right|_{r=1, s=-1, t=2}\) is 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is an essential concept in calculus, particularly when dealing with functions of multiple variables. It helps us find the derivative of composite functions. In the context of multivariable calculus, the chain rule allows us to differentiate a function even when it is expressed in terms of other dependent variables.

When applying the chain rule, you proceed by differentiating the outer function with respect to the inner function, which may involve multiple steps:
  • Identify inner functions and their dependencies.
  • Compute the derivative of the outer function with respect to the inner functions.
  • Multiply by the derivatives of the inner functions with respect to your variable of interest.
Using the chain rule in our problem, even if indirectly, involves substituting expressions for variables like replacing \(x\) with \(r + s + t\) and \(y\) with \(rst\) in the function \(z = xy + x + y\). This setup ensures we accommodate all dependencies between variables, allowing accurate computation of partial derivatives.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to more than one variable, bringing in new concepts such as partial derivatives, multiple integrals, and more. One of the key challenges addressed by multivariable calculus is how variables interact and contribute to the overall function behavior.

In the given exercise, we see a function \(z = xy + x + y\) where both \(x\) and \(y\) are in terms of multiple variables \(r\), \(s\), and \(t\). This complex dependency is typical in multivariable contexts:
  • Variables influence each other directly or indirectly, as seen in the implicit relationships among \(x = r + s + t\) and \(y = rst\).
  • Functions like \(z\) depend on multiple interrelated variables, requiring special techniques for differentiation.
  • Partial derivatives, as used here, help in finding the rate of change of a function with respect to one variable at a time, keeping others constant.
By understanding these interactions through multivariable calculus, we can solve for derivatives that reveal how one variable, such as \(s\), influences the function \(z\) independently of the others.
Implicit Differentiation
Implicit differentiation is a technique used when a function is not explicitly solved for one variable in terms of others. This approach is handy for expressions where explicitly solving for one variable would be difficult or impossible.

Although our function \(z = xy + x + y\) is not directly requiring implicit differentiation, it involves expressions where certain dependencies could make implicit differentiation relevant. Here are the general steps:
  • Differentiate each term with respect to the desired variable, treating all other terms as functions of that variable.
  • Apply the chain rule where necessary for any composite terms.
  • Collect all derivatives and solve accordingly.
The solution for finding \(\frac{\partial z}{\partial s}\) involves treating certain terms as constants and others as functions, similar to strategies used in implicit differentiation. This way, we systematically isolate the effect of a single variable, leading to the accurate evaluation of partial derivatives in complex situations.

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