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Chapter 5: Problem 6

Find \(h_{y}\) and \(h_{z}\) if $$ h(y, z)=g(x(y, z), y, z, w(y, z)) $$

Short Answer

Expert verified
Question: Determine the partial derivatives \(h_{y}\) and \(h_{z}\) of the function \(h(y, z)\), which is defined as \(g(x(y, z), y, z, w(y, z))\). Answer: To find the partial derivatives \(h_{y}\) and \(h_{z}\), we use the chain rule as follows: - For \(h_{y}\): $$ h_{y} = \frac{\partial g}{\partial x} \cdot \frac{\partial x(y, z)}{\partial y} + \frac{\partial g}{\partial y} + \frac{\partial g}{\partial z} \cdot \frac{\partial z}{\partial y} + \frac{\partial g}{\partial w} \cdot \frac{\partial w(y, z)}{\partial y} $$ - For \(h_{z}\): $$ h_{z} = \frac{\partial g}{\partial x} \cdot \frac{\partial x(y, z)}{\partial z} + \frac{\partial g}{\partial y} \cdot \frac{\partial y}{\partial z} + \frac{\partial g}{\partial z} + \frac{\partial g}{\partial w} \cdot \frac{\partial w(y, z)}{\partial z} $$ The specific expressions for \(h_{y}\) and \(h_{z}\) will depend on the given functions \(g\), \(x(y, z)\), and \(w(y, z)\) and their respective partial derivatives.

Step by step solution

01

Understand the function

The given function \(h(y, z)\) depends on \(y\) and \(z\). The function \(h\) is defined as \(g(x(y, z), y, z, w(y, z))\). Therefore, we need to apply the chain rule to find the partial derivatives \(h_{y}\) and \(h_{z}\). 2. Apply the chain rule to find \(h_{y}\)
02

Compute \(h_{y}\) using the chain rule

To find \(h_{y}\), we will differentiate \(h(y,z)\) with respect to \(y\), applying the chain rule: $$ h_{y} = \frac{\partial g}{\partial x} \cdot \frac{\partial x(y, z)}{\partial y} + \frac{\partial g}{\partial y} + \frac{\partial g}{\partial z} \cdot \frac{\partial z}{\partial y} + \frac{\partial g}{\partial w} \cdot \frac{\partial w(y, z)}{\partial y} $$ Here, we have used the chain rule, and we need to find the individual partial derivatives. 3. Apply the chain rule to find \(h_{z}\)
03

Compute \(h_{z}\) using the chain rule

To find \(h_{z}\), we will differentiate \(h(y,z)\) with respect to \(z\), applying the chain rule: $$ h_{z} = \frac{\partial g}{\partial x} \cdot \frac{\partial x(y, z)}{\partial z} + \frac{\partial g}{\partial y} \cdot \frac{\partial y}{\partial z} + \frac{\partial g}{\partial z} + \frac{\partial g}{\partial w} \cdot \frac{\partial w(y, z)}{\partial z} $$ Just like we did for \(h_{y}\), here we have used the chain rule, and we need to find the individual partial derivatives. 4. Identify the required partial derivatives
04

Partial derivatives required for \(h_{y}\) and \(h_{z}\)

To find \(h_{y}\) and \(h_{z}\), we need to compute the following partial derivatives: - \(\frac{\partial x(y, z)}{\partial y}\) and \(\frac{\partial x(y, z)}{\partial z}\) - \(\frac{\partial y}{\partial z}\) - \(\frac{\partial w(y, z)}{\partial y}\) and \(\frac{\partial w(y, z)}{\partial z}\) - \(\frac{\partial g}{\partial x}\), \(\frac{\partial g}{\partial y}\), \(\frac{\partial g}{\partial z}\), and \(\frac{\partial g}{\partial w}\). These partial derivatives will be given or calculated based on the specific functions \(g\), \(x(y, z)\), and \(w(y, z)\). 5. Calculate \(h_{y}\) and \(h_{z}\)
05

Compute \(h_{y}\) and \(h_{z}\) based on the partial derivatives

Use the calculated partial derivatives and the chain rule expressions for \(h_{y}\) and \(h_{z}\) from steps 2 and 3 to compute the final expressions for \(h_{y}\) and \(h_{z}\).

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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 a fundamental tool in calculus, particularly useful when dealing with functions that are compositions of other functions. In the context of multivariable calculus, where functions often depend on several variables, the chain rule enables us to calculate the derivative of a composite function with respect to any of its variables.

Considering the exercise at hand, we need to find the partial derivatives of the function h with respect to y and z, where h is expressed in terms of yet another function g, which itself is dependent on y, z, as well as other functions of y and z. Consequently, applying the chain rule correctly necessitates differentiating g with respect to each of its arguments鈥攅ven those that are functions鈥攁nd then multiplying by the derivative of those argument functions with respect to y or z. It's like untangling a knot, where each part of the knot is carefully loosened to avoid tangling other parts.

Significance of Mastering the Chain Rule

When mastered, the chain rule provides precision and clarity to differentiation. Without proper use of the chain rule, calculations can be incorrect or incomplete, leading to misunderstandings in more advanced topics such as optimization and modeling in various fields of engineering and science.
Real Analysis
Real analysis is the branch of mathematics that deals with the rigorous study of real numbers, sequences, functions, and their derivatives and integrals. It lays the foundation for all of calculus and, by extension, much of modern mathematics and physics. In real analysis, we focus on proving the concepts and rules we often use intuitively in calculus, such as limits, continuity, and differentiability.

In solving the exercise, we encounter a situation where real analysis plays a key role: the computation of partial derivatives. Understanding the limit process that defines derivatives enables us to handle more complex cases, such as functions of multiple variables, with confidence. It's akin to knowing the rules of grammar in a language; the better you know them, the more complex and expressive sentences you can construct.

Why Real Analysis Matters

A deep understanding of real analysis is essential for any advanced work in mathematics, including the proper application of the chain rule in multivariable calculus. This is because it ensures the logical structure underlying the procedures鈥攆or instance, it assures us that the limit definitions are satisfied, and that functions behave as expected when subjected to the rigors of differentiation and integration.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus (like derivatives and integrals) to functions of multiple variables. Unlike single-variable calculus where the function's output depends on one input, multivariable calculus deals with functions whose outputs can be influenced by two or more inputs. This is particularly important in fields like physics, economics, and engineering where many factors often affect outcomes simultaneously.

In the given exercise, we are dealing with the partial derivatives of a multivariable function, which measure how the function changes as each input is varied while holding the others constant. Computing these derivatives allows us to understand the sensitivity of the function's output to changes in any one of the independent variables.

Real-World Applications of Multivariable Calculus

Proficiency in multivariable calculus is crucial for anyone aiming to model complex systems. Whether it's predicting weather changes, optimizing resource allocation in businesses, or designing intricate engineering systems, the ability to dissect these systems into their constituent relationships via partial derivatives is a powerful skill that stems from multivariable calculus.

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Most popular questions from this chapter

Let \(r_{1}, r_{2}, \ldots, r_{n}\) be nonnegative integers such that $$r_{1}+r_{2}+\cdots+r_{n}=r \geq 0$$ (a) Show that $$\left(z_{1}+z_{2}+\cdots+z_{n}\right)^{r}=\sum_{r} \frac{r !}{r_{1} ! r_{2} ! \cdots r_{n} !} z_{1}^{r_{1}} z_{2}^{r_{2}} \cdots z_{n}^{r_{n}}$$ where \(\sum_{r}\) denotes summation over all \(n\) -tuples \(\left(r_{1}, r_{2}, \ldots, r_{n}\right)\) that satisfy the stated conditions. HINT: This is obvious if \(n=1,\) and it follows from Exercise 1.2 .19 if \(n=2 .\) Use induction on \(n\). (b) Show that there are $$\frac{r !}{r_{1} ! r_{2} ! \cdots r_{n} !}$$ ordered \(n\) -tuples of integers \(\left(i_{1}, i_{2}, \ldots, i_{n}\right)\) that contain \(r_{1}\) ones, \(r_{2}\) twos, ... and \(r_{n} n\) 's. (c) Let \(f\) be a function of \(\left(x_{1}, x_{2}, \ldots, x_{n}\right)\). Show that there are $$\frac{r !}{r_{1} ! r_{2} ! \cdots r_{n} !}$$ partial derivatives \(f_{x_{i} 1} x_{i_{2}} \cdots x_{i_{r}}\) that involve differentiation \(r_{i}\) times with respect to \(x_{i},\) for \(i=1,2, \ldots, n\).

Let \(h(\mathbf{U})=f(\mathbf{G}(\mathbf{U}))\) and find \(d_{\mathbf{U}_{0}} h\) by Theorem 5.4.3, and then by writing \(h\) explicitly as a function of \(\mathbf{U}\). (a) \(\begin{aligned} f(x, y) &=3 x^{2}+4 x y^{2}+3 x \\ g_{1}(u, v) &=v e^{u+v-1}, \\ g_{2}(u, v) &=e^{-u+v-1} \end{aligned} \quad\left(u_{0}, v_{0}\right)=(0,1)\) (b) \(\begin{aligned} f(x, y, z) &=e^{-(x+y+z)} \\ g_{1}(u, v, w) &=\log u-\log v+\log w \\ g_{2}(u, v, w) &=-2 \log u-3 \log w \\ g_{3}(u, v, w) &=\log u+\log v+2 \log w \end{aligned}\) \(\left(u_{0}, v_{0}, w_{0}\right)=(1,1,1)\) (c) \(\begin{aligned} f(x, y) &=(x+y)^{2}, \\ g_{1}(u, v) &=u \cos v, \quad\left(u_{0}, v_{0}\right)=(3, \pi / 2) \\ g_{2}(u, v) &=u \sin v, \end{aligned}\) (d) \(\begin{aligned} f(x, y, z) &=x^{2}+y^{2}+z^{2} \\ g_{1}(u, v, w) &=u \cos v \sin w \\ g_{2}(u, v, w) &=u \cos v \cos w, \\ g_{3}(u, v, w) &=u \sin v \end{aligned}\) \(\left(u_{0}, v_{0}\right)=(3, \pi / 2)\)

Let \(D_{1}\) and \(D_{2}\) be compact subsets of \(\mathbb{R}^{n}\). Show that $$ D=\left\\{(\mathbf{X}, \mathbf{Y}) \mid \mathbf{X} \in D_{1}, \mathbf{Y} \in D_{2}\right\\} $$ is a compact subset of \(\mathbf{R}^{2 n}\).

Show that the function $$f(x, y)=\left\\{\begin{array}{ll} \frac{x^{2} y}{x^{6}+2 y^{2}}, & (x, y) \neq(0,0), \\ 0, & (x, y)=(0,0), \end{array}\right.$$ has a directional derivative in the direction of an arbitrary unit vector \(\Phi\) at \((0,0),\) but \(f\) is not continuous at (0,0) .

If \(h(r, \theta)=f(r \cos \theta, r \sin \theta),\) show that $$ f_{x x}+f_{y y}=h_{r r}+\frac{1}{r} h_{r}+\frac{1}{r^{2}} h_{\theta \theta} $$ HINT: Rewrite the defining equation as \(f(x, y)=h(r(x, y), \theta(x, y)),\) with \(r(x, y)=\) \(\sqrt{x^{2}+y^{2}}\) and \(\theta(x, y)=\tan ^{-1}(y / x),\) and differentiate with respect to \(x\) and \(y .\)
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