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In Exercises \(23-34,\) find \(f_{x}, f_{y},\) and \(f_{z}\) $$f(x, y, z)=\tanh (x+2 y+3 z)$$

Short Answer

Expert verified
\(f_x = \text{sech}^2(u), f_y = 2 \cdot \text{sech}^2(u), f_z = 3 \cdot \text{sech}^2(u)\), where \(u = x+2y+3z\).

Step by step solution

01

Recognize the Structure

Given the function \(f(x, y, z) = \tanh(x + 2y + 3z)\), we need to find the partial derivatives \(f_x\), \(f_y\), and \(f_z\). Each partial derivative will involve treating the other variables as constants.
02

Calculate \(f_x\)

To find \(f_x\), differentiate \(f(x, y, z)\) with respect to \(x\) using the chain rule. Since the derivative of \(\tanh(u)\) is \(\text{sech}^2(u)\), where \(u = x + 2y + 3z\), we get \[f_x = \text{sech}^2(x + 2y + 3z) \cdot 1 = \text{sech}^2(x + 2y + 3z)\.\]
03

Calculate \(f_y\)

To find \(f_y\), we differentiate \(f(x, y, z)\) with respect to \(y\). Apply the chain rule: \[f_y = \text{sech}^2(x + 2y + 3z) \cdot 2 = 2 \cdot \text{sech}^2(x + 2y + 3z)\.\]
04

Calculate \(f_z\)

To find \(f_z\), differentiate \(f(x, y, z)\) with respect to \(z\). Again, apply the chain rule: \[f_z = \text{sech}^2(x + 2y + 3z) \cdot 3 = 3 \cdot \text{sech}^2(x + 2y + 3z)\.\]

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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 used to differentiate composite functions. When dealing with multiple variables, this rule helps us find how changes in one variable affect a function, considering that other variables might also be changing.

To use the chain rule, imagine breaking a complex function into simpler parts, then finding derivatives of each part. For a function like \(f(x, y, z) = \tanh(x + 2y + 3z)\), which involves a composition of the tangent hyperbolic function and a linear combination of variables, the chain rule simplifies the process of differentiation.

The basic idea is:
  • Differentiate the outer function: \(\tanh(u)\) gives \(\text{sech}^2(u)\).
  • Then, multiply it by the derivative of the inner function with respect to the variable of interest.
For instance, to find \(f_x\), differentiate \(u = x + 2y + 3z\) with respect to \(x\), which equals 1, resulting in \(f_x = \text{sech}^2(u) \cdot 1\). For \(f_y\) and \(f_z\), follow similarly by differentiating \(u\) with respect to \(y\) and \(z\), leading to multipliers of 2 and 3, respectively.
Hyperbolic Functions
Hyperbolic functions, like \(\tanh, \sinh,\) and \(\cosh\), are analogs of trigonometric functions but for a hyperbola instead of a circle. The \(\tanh\) function is particularly interesting in calculus due to its unique properties and symmetry.

The \(\tanh(u)\) function, or hyperbolic tangent, is defined as:
  • \(\tanh(u) = \frac{\sinh(u)}{\cosh(u)}\)
  • Where \(\sinh(u) = \frac{e^u - e^{-u}}{2}\) and \(\cosh(u) = \frac{e^u + e^{-u}}{2}\)
This gives \(\tanh(u)\) similar properties to \(\sin(u)\), but while the sine function oscillates between -1 and 1, \(\tanh(u)\) approaches these limits asymptotically.

When taking derivatives, the nice property is that \(\frac{d}{du}\tanh(u) = \text{sech}^2(u)\), which means differentiating \(\tanh(u)\) results in another hyperbolic function, the hyperbolic secant squared, making these operations very smooth.
Multivariable Calculus
Multivariable calculus expands the principles of calculus to functions involving more than one variable. It allows us to study how functions change as each input variable changes, while holding others constant.

In the problem \(f(x, y, z) = \tanh(x + 2y + 3z)\), the task is to find partial derivatives \(f_x, f_y,\) and \(f_z\). Partial derivatives show the rate of change of a function with respect to one variable, ignoring the effect of changes in other variables.
  • For \(f_x\), treat \(y\) and \(z\) as constants and differentiate with respect to \(x\).
  • Similarly, for \(f_y\) and \(f_z\), treat the other variables as constants.
This method of holding variables static while differentiating one at a time simplifies complex relations and aids in constructing tools like gradients and tangent planes. Multivariable functions often require partial derivatives for modeling real-world phenomena such as thermodynamics, economics, and more, where multiple factors influence outcomes simultaneously.

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Most popular questions from this chapter

Use a CAS to plot the implicitly defined level surfaces in Exercises \(73-76 .\) $$ 4 \ln \left(x^{2}+y^{2}+z^{2}\right)=1 $$

If a function \(f(x, y)\) has continuous second partial derivatives throughout an open region \(R,\) must the first-order partial derivatives of \(f\) be continuous on \(R ?\) Give reasons for your answer.

Let $$f(x, y)=\left\\{\begin{array}{ll}{x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}},} & {\text { if }(x, y) \neq 0} \\ {0,} & {\text { if }(x, y)=0}\end{array}\right.$$ a. Show that \(\frac{\partial f}{\partial y}(x, 0)=x\) for all \(x,\) and \(\frac{\partial f}{\partial x}(0, y)=-y\) for all \(y\) b. Show that \(\frac{\partial^{2} f}{\partial y \partial x}(0,0) \neq \frac{\partial^{2} f}{\partial x \partial y}(0,0)\) The graph of \(f\) is shown on page 788 . The three-dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}=0$$ is satisfied by steady-state temperature distributions \(T=f(x, y, z)\) in space, by gravitational potentials, and by electrostatic potentials. The two-dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0$$ obtained by dropping the \(\partial^{2} f / \partial z^{2}\) term from the previous equation, describes potentials and steady-state temperature distributions in a plane (see the accompanying figure). The plane (a) may be treated as a thin slice of the solid (b) perpendicular to the \(z\) -axis.

Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I,\) you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$ \begin{array}{l}{x=u \cos v, \quad y=u \sin v, \quad z=u, \quad 0 \leq u \leq 2} \\ {0 \leq v \leq 2 \pi}\end{array} $$

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=\left\\{\begin{array}{ll}{x^{5} \ln \left(x^{2}+y^{2}\right),} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$ $$-2 \leq x \leq 2, \quad-2 \leq y \leq 2$$

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