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Chapter 14: Problem 46

$$ \begin{array}{l}{\text { Assume that } z=\ln (f(w)), w=g(x, y), \quad x=\sqrt{r-s}, \quad \text { and }} \\ {y=r^{2} s . \quad \text { If } \quad g_{x}(2,-9)=-1, \quad g_{y}(2,-9)=3, \quad f^{\prime}(-2)=2} \\ {f(-2)=5, \quad \text { and } \quad g(2,-9)=-2, \quad \text { find } \quad\left.\frac{\partial z}{\partial r}\right|_{r=3, s=-1} \text { and }} \\\ {\left.\frac{\partial z}{\partial s}\right|_{r=3, s=-1}}\end{array} $$

Short Answer

Expert verified
\( \frac{\partial z}{\partial r} = -\frac{37}{10} \), \( \frac{\partial z}{\partial s} = \frac{109}{10} \) at \( r = 3, s = -1 \).

Step by step solution

01

Identify the Problem

We need to find \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial s} \) where \( z = \ln(f(w)) \), \( w = g(x, y) \), \( x = \sqrt{r-s} \), and \( y = r^2 s \). We are given several function values and partial derivatives.
02

Use Chain Rule for Partial Derivatives

To find \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial s} \), we need to use the chain rule for multivariable functions. Specifically, we have:\[ \frac{\partial z}{\partial r} = \frac{\partial z}{\partial w} \cdot \frac{\partial w}{\partial r} \]\[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial w} \cdot \frac{\partial w}{\partial s} \]
03

Compute \( \frac{\partial z}{\partial w} \)

Since \( z = \ln(f(w)) \), by differentiating with respect to \( w \), we get:\[ \frac{\partial z}{\partial w} = \frac{1}{f(w)} \cdot f'(w) \] Given that \( f(-2) = 5 \) and \( f'(-2) = 2 \), and since \( w = -2 \) when \( r = 3 \) and \( s = -1 \), we have:\[ \frac{\partial z}{\partial w} = \frac{1}{5} \cdot 2 = \frac{2}{5} \]
04

Find \( \frac{\partial w}{\partial r} \) and \( \frac{\partial w}{\partial s} \)

For \( w = g(x, y) \):- \( x = \sqrt{r-s} \), thus \( \frac{\partial x}{\partial r} = \frac{1}{2\sqrt{r-s}} \) and \( \frac{\partial x}{\partial s} = -\frac{1}{2\sqrt{r-s}} \).- \( y = r^2s \), thus \( \frac{\partial y}{\partial r} = 2rs \) and \( \frac{\partial y}{\partial s} = r^2 \).Using the chain rule:\[ \frac{\partial w}{\partial r} = g_x \frac{\partial x}{\partial r} + g_y \frac{\partial y}{\partial r} \]\[ \frac{\partial w}{\partial s} = g_x \frac{\partial x}{\partial s} + g_y \frac{\partial y}{\partial s} \]
05

Compute \( \frac{\partial w}{\partial r} \)

Calculate \( \frac{\partial w}{\partial r} \) at \( r = 3 \), \( s = -1 \) using given values:- \( x = \sqrt{3 + 1} = 2 \), \( y = 3^2(-1) = -9 \).- \( g_x(2, -9) = -1 \), \( g_y(2, -9) = 3 \).\[ \frac{\partial w}{\partial r} = (-1) \cdot \frac{1}{4} + 3 \cdot (-3) = -\frac{1}{4} - 9 = -\frac{37}{4} \]
06

Compute \( \frac{\partial w}{\partial s} \)

Calculate \( \frac{\partial w}{\partial s} \) at \( r = 3 \), \( s = -1 \):- \( \frac{\partial x}{\partial s} = -\frac{1}{4} \), \( \frac{\partial y}{\partial s} = 9 \).\[ \frac{\partial w}{\partial s} = (-1) \cdot -\frac{1}{4} + 3 \cdot 9 = \frac{1}{4} + 27 = \frac{109}{4} \]
07

Calculate \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial s} \)

Substitute values into the chain rule formulas:\[ \frac{\partial z}{\partial r} = \frac{2}{5} \cdot \left( -\frac{37}{4} \right) = -\frac{37}{10} \]\[ \frac{\partial z}{\partial s} = \frac{2}{5} \cdot \frac{109}{4} = \frac{109}{10} \]

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Key Concepts

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

Chain Rule
In multivariable calculus, the chain rule is essential when dealing with functions of multiple variables. If a function depends indirectly on a variable through other intermediate variables, the chain rule allows us to differentiate it. For example, if we have a function \( z \) that depends on another function \( w \), and \( w \) itself depends on variables \( x \) and \( y \), the total derivative of \( z \) with respect to any other variables can be found using the chain rule.
  • The chain rule for multivariable functions states that the derivative of \( z \) with respect to \( r \) can be expressed as: \[ \frac{\partial z}{\partial r} = \frac{\partial z}{\partial w} \cdot \frac{\partial w}{\partial r} \]
  • Similarly, \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial w} \cdot \frac{\partial w}{\partial s} \).
Understanding this approach involves calculating partial derivatives at each stage and then multiplying them accordingly to get the desired derivative.
Partial Derivatives
Partial derivatives are derivatives of functions with several variables, where we choose to differentiate with respect to one variable while treating others as constants. This approach is crucial when dealing with functions of more than one variable. In our example, we need to find partial derivatives like \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial s} \).
  • To find \( \frac{\partial w}{\partial r} \), we use the derivatives of \( x \) and \( y \) since \( w = g(x, y) \).
  • This yields: \( \frac{\partial w}{\partial r} = g_x \cdot \frac{\partial x}{\partial r} + g_y \cdot \frac{\partial y}{\partial r} \) for function \( g(x, y) \).
Partial derivatives help in tracing how changes in one variable affect the function, keeping others constant.
Multivariable Functions
Multivariable functions are functions that depend on more than one variable. They appear in calculus to describe real-world phenomena where multiple factors influence outcomes.
  • In this problem, \( z = \ln(f(w)) \) is a function influenced indirectly by \( r \) and \( s \) through intermediate functions like \( w = g(x, y) \).
  • Thus, understanding how \( w \), \( x \), and \( y \) contribute to \( z \) ultimately describes how the function behaves.
This stacking of functions makes differentiation complex but possible with the chain rule, helping us determine how one variable affects another through multiple layers of dependencies.
Logarithmic Differentiation
Logarithmic differentiation comes into play when dealing with exponential or logarithmic functions. It can simplify the process of differentiation, particularly when functions take a logarithmic form as in \( z = \ln(f(w)) \).
  • The partial derivative of \( z \) concerning \( w \) is given by: \[ \frac{\partial z}{\partial w} = \frac{1}{f(w)} \cdot f'(w) \]
  • This equation simplifies the differentiation process, reducing it to multiplication of functions when compared to its traditional form.
Logarithmic differentiation is notably helpful for functions that are products or quotients, often leading to simpler derivatives. In this case, it lets us factor out the logarithmic function's complexity early on.

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Most popular questions from this chapter

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