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Chapter 12: Problem 3

Find the first partial derivatives of the function. \(f(x, y)=2 x+3 y+5\)

Short Answer

Expert verified
The first partial derivatives of the function \(f(x, y) = 2x + 3y + 5\) are: \(f_x(x, y) = 2\) \(f_y(x, y) = 3\)

Step by step solution

01

Compute the Partial Derivative with respect to x (f_x)

Since y is considered constant while differentiating with respect to x, we can differentiate the function f(x, y) = 2x + 3y + 5 directly with respect to x: f_x(x, y) = \(\frac{\partial}{\partial x}\)(2x + 3y + 5) f_x(x, y) = 2 Here, the partial derivative f_x is simply a constant, 2.
02

Compute the Partial Derivative with respect to y (f_y)

Since x is considered constant while differentiating with respect to y, we can differentiate the function f(x, y) = 2x + 3y + 5 directly with respect to y: f_y(x, y) = \(\frac{\partial}{\partial y}\)(2x + 3y + 5) f_y(x, y) = 3 Here, the partial derivative f_y is simply a constant, 3.
03

Final Answer

The first partial derivatives of the function f(x, y) = 2x + 3y + 5 are: f_x(x, y) = 2 f_y(x, y) = 3

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of single-variable calculus to functions of several variables. While single-variable calculus deals with functions of just one variable, multivariable calculus opens the door to exploring functions that depend on two, three, or even more variables. This branch of mathematics is crucial for modeling real-world scenarios where multiple factors come into play.

**Key Features of Multivariable Calculus:**
  • Functions of multiple variables: For instance, the function in the exercise, \( f(x, y) = 2x + 3y + 5 \), depends both on \(x\) and \(y\).
  • Visualization: It allows us to visualize functions in higher dimensions, using concepts like surfaces and contours.
  • Applications: Useful in physics for calculating gravitational fields, in economics for optimizing production, and much more.
Grasping multivariable calculus can enrich your understanding of how complex systems operate across different domains.
Differentiation
Differentiation is a fundamental concept in calculus used to understand how a function changes as its variables change. In the context of multivariable functions, differentiation manifests through partial derivatives, which allow us to investigate the rate of change with respect to one specific variable at a time.

**Differentiation in Multivariable Functions**
  • Procedure: Consider the effect of changing just one variable at a time, treating all other variables as constants.
  • Goal: Find the 'slope' or rate of change of the function along that axis or direction.
  • Relevance: Helps in finding tangent planes and optimizing functions over several variables.
Using differentiation, we can better predict the behavior of complex multivariable functions in various applications like engineering and economics.
Partial Derivative Rules
Partial derivative rules are specific guidelines that help us perform differentiation with respect to one variable while considering others constant. They form an essential toolset to understand multivariable calculus, particularly useful in simplifying the process of finding derivatives like the ones in the original exercise.

**Basic Rules for Partial Derivatives:**
  • Treat other variables as constants: For \(f(x, y) = 2x + 3y + 5\), during \(\frac{\partial}{\partial x}\), treat \(y\) as a constant.
  • Linearity rule: Derivatives of sums and constants are computed separately and then combined together.
  • Constant rule: If a term does not contain the variable in question, its derivative is zero.
Understanding and applying these rules make it simpler to handle functions with multiple variables, enhancing proficiency in both academic and practical scenarios.

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Most popular questions from this chapter

Find the first partial derivatives of the function. \(f(s, t)=\left(s^{2}-s t+t^{2}\right)^{3}\)

The monthly profit (in dollars) of Bond and Barker Department Store depends on the level of inventory \(x\) (in thousands of dollars) and the floor space \(y\) (in thousands of square feet) available for display of the merchandise, as given by the equation $$\begin{aligned}P(x, y)=&-0.02 x^{2}-15 y^{2}+x y \\ &+39 x+25 y-20,000\end{aligned}$$ Compute \(\partial P / \partial x\) and \(\partial P / \partial y\) when \(x=4000\) and \(y=150\). Interpret your results. Repeat with \(x=5000\) and \(y=150\).

The total daily revenue (in dollars) that Weston Publishing realizes in publishing and selling its English-language dictionaries is given by $$\begin{aligned}R(x, y)=&-0.005 x^{2}-0.003 y^{2}-0.002 x y \\\&+20 x+15 y\end{aligned}$$ where \(x\) denotes the number of deluxe copies and \(y\) denotes the number of standard copies published and sold daily. The total daily cost of publishing these dictionaries is given by $$C(x, y)=6 x+3 y+200$$ dollars. Determine how many deluxe copies and how many standard copies Weston should publish each day to maximize its profits. What is the maximum profit realizable?

Let \(f(x, y)=9-x^{2}+x y-2 y^{2}\). a. Find \(f_{x}(1,2)\) and \(f_{y}(1,2)\). b. Interpret the numbers in part (a) as slopes. c. Interpret the numbers in part (a) as rates of change.

The monthly payment that amortizes a loan of \(A\) dollars in \(t\) yr when the interest rate is \(r\) per year, compounded monthly, is given by $$P=f(A, r, t)=\frac{A r}{12\left[1-\left(1+\frac{r}{12}\right)^{-12 t}\right]}$$ a. What is the monthly payment for a home mortgage of $$\$ 300,000$$ that will be amortized over \(30 \mathrm{yr}\) with an interest rate of \(6% /\) year? An interest rate of \(8 \% /\) year? b. Find the monthly payment for a home mortgage of $$\$ 300,000$$ that will be amortized over 20 yr with an interest rate of \(8 \% /\) year.
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