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Chapter 9: Problem 29

Find the first partial derivatives of the function. $$f(x, y, z, t)=x y z^{2} \tan (y t)$$

Short Answer

Expert verified
Partial derivatives: \( \frac{\partial f}{\partial x} = yz^2 \tan(yt) \), \( \frac{\partial f}{\partial y} = xz^2 \tan(yt) + xytz^2 \sec^2(yt) \), \( \frac{\partial f}{\partial z} = 2xyz \tan(yt) \), and \( \frac{\partial f}{\partial t} = x y^2 z^2 \sec^2(yt) \).

Step by step solution

01

Understand Partial Derivatives

Partial derivatives involve taking the derivative of a function with respect to one variable while keeping other variables constant. In this problem, the function \( f(x, y, z, t) = x y z^2 \tan(yt) \) requires finding the derivatives with respect to each variable: \(x\), \(y\), \(z\), and \(t\).
02

Differentiate with respect to x

To find \( \frac{\partial f}{\partial x} \), treat \(y\), \(z\), and \(t\) as constants. Differentiating \( x \cdot y \cdot z^2 \cdot \tan(yt) \) with respect to \(x\) gives:\[ \frac{\partial f}{\partial x} = yz^2 \tan(yt) \]
03

Differentiate with respect to y

To find \( \frac{\partial f}{\partial y} \), treat \(x\), \(z\), and \(t\) as constants. Use the product rule as the function involves both \(y\) and \(\tan(yt)\):\[ \frac{\partial f}{\partial y} = xz^2 \tan(yt) + xytz^2 \sec^2(yt) \]
04

Differentiate with respect to z

To find \( \frac{\partial f}{\partial z} \), treat \(x\), \(y\), and \(t\) as constants. Differentiating \(x y z^2 \tan(yt)\) with respect to \(z\) gives:\[ \frac{\partial f}{\partial z} = 2xyz \tan(yt) \]
05

Differentiate with respect to t

To find \( \frac{\partial f}{\partial t} \), treat \(x\), \(y\), and \(z\) as constants. Use the chain rule on \(\tan(yt)\), which becomes \(y \sec^2(yt)\) when differentiated with respect to \(t\):\[ \frac{\partial f}{\partial t} = x y^2 z^2 \sec^2(yt) \]

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Key Concepts

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

Function Differentiation
Differentiation is a fundamental concept in calculus that helps to determine the rate at which a function changes. When we talk about *partial derivatives*, we are looking at functions that have more than one variable, and we're interested in how the function changes with respect to just one of those variables.
  • When we find the derivative of a function with a single variable, it's straightforward because there's only one rate of change.
  • In the case of multivariable functions, like the one in our exercise, we have to specify the variable of interest while keeping the others constant.
  • This is why we call it a "partial" derivative - because it only applies to part of the function.
So, in this context, *differentiation* means finding how changes in one particular variable, say \(x\), affect the value of the function, while assuming that the other variables \(y, z,\) and \(t\) remain unchanged.
Multivariable Calculus
Multivariable calculus is an extension of standard calculus to functions of several variables. Instead of just looking at one-dimensional lines, we consider more complex surfaces and shapes.
  • When dealing with functions of multiple variables, calculus allows us to explore how each variable influences the function's behavior.
  • For instance, the function \( f(x, y, z, t) = x y z^2 \tan(yt) \) involves four variables, each potentially affecting the output differently.
  • By using partial derivatives, we can examine how the function behaves when viewed along different slices or dimensions.
This enhances our understanding of dynamic systems where several variables interact, such as in physics or economics. Multivariable calculus employs tools like partial derivatives, gradients, and integrals to explore these complex interdependencies.
Product Rule
The product rule is a technique used in calculus for finding the derivative of a product of two or more functions. When functions are multiplied together, their derivatives are not simply the product of the individual derivatives.
  • For the exercise function \( f(x, y, z, t) = x y z^2 \tan(yt) \), applying the product rule is key when differentiating with respect to \(y\).
  • The product rule states that the derivative of \( u \cdot v \) is \( u'v + uv' \), where \( u \) and \( v \) are functions of the variable you're differentiating with respect to.
  • In this exercise, \( y \) and \( \tan(yt) \) are treated as functions of \( y \), requiring the use of the product rule to find \( \frac{\partial f}{\partial y} \).
This rule is incredibly useful as it breaks down complex products into simpler, more manageable parts.
Chain Rule
The chain rule is a critical rule in calculus used to differentiate compositions of functions. It allows us to differentiate functions that are nested within other functions.
  • For example, consider differentiating \( \tan(yt) \) with respect to \( t \). Here, \( \tan \) is a function nested within itself a linear transformation of \( yt \).
  • The chain rule states: If you have a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).
  • In the exercise, the inner function \( g(t) \) is \( yt \), and the outer function is \( \tan(g(t)) \). The derivative involves both \( \sec^2(yt) \), the derivative of the outer function, and \( y \), the derivative of the inner function w.r.t. \( t \).
By using the chain rule, we can effectively "chain" together derivatives to get the derivative of the composite function, simplifying complex differentiations.

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Most popular questions from this chapter

Determine the set of points at which the function is continuous. \(r(x, y)=\arctan (x+\sqrt{y})\)

Find the first partial derivatives of the function. $$u=t e^{w / t}$$

A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units/m \(^{2}\) per day, the north and south walls at a rate of 8 units/m \(^{2}\) per day, the floor at a rate of 1 unit/m \(^{2}\) per day, and the roof at a rate of 5 units/m \(^{2}\) per day. Each wall mustbe at least 30 \(\mathrm{m}\) long, the height must be at least \(4 \mathrm{m},\) and the volume must be exactly 4000 \(\mathrm{m}^{3} .\) (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. (b) Find the dimensions that minimize heat loss. (Check both the critical points and the points on the boundary of the domain.) (c) Could you design a building with even less heat loss if the restrictions on the lengths of the walls were removed?

Find all the second partial derivatives. $$v=e^{x e^{y}}$$

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