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Chapter 14: Problem 39

Find the indicated partial derivatives. $$f(x, y)=\ln \left(x+\sqrt{x^{2}+y^{2}}\right) ; \quad f_{x}(3,4)$$

Short Answer

Expert verified
The partial derivative \( f_x(3, 4) = \frac{1}{5} \).

Step by step solution

01

Understanding the Function

We are given the function \( f(x, y) = \ln \left( x + \sqrt{x^2 + y^2} \right) \). We need to find the partial derivative of this function with respect to \( x \), which is denoted as \( f_x(x, y) \). This partial derivative will be evaluated at the point \((3,4)\).
02

Applying the Chain Rule

To find \( f_x(x, y) \), we will use the chain rule. The chain rule is particularly useful here because \( f(x, y) \) is a composition of the natural logarithm and the inside function \( g(x, y) = x + \sqrt{x^2 + y^2} \). That means \( f(x, y) = \ln(g(x, y)) \). Hence, the derivative is \( f_x = \frac{1}{g(x, y)} \cdot g_x(x, y) \).
03

Differentiating the Inside Function

Now, we need to calculate \( g_x(x, y) \). The function \( g(x, y) = x + \sqrt{x^2 + y^2} \) has two parts. The derivative of \( x \) with respect to \( x \) is 1. For \( \sqrt{x^2 + y^2} \), use the chain rule: the derivative is \( \frac{x}{\sqrt{x^2 + y^2}} \). So, \( g_x(x, y) = 1 + \frac{x}{\sqrt{x^2 + y^2}} \).
04

Finding the Partial Derivative

Substitute \( g(x, y) = x + \sqrt{x^2 + y^2} \) and \( g_x(x, y) = 1 + \frac{x}{\sqrt{x^2 + y^2}} \) back into the expression for \( f_x \) from Step 2: \[ f_x(x, y) = \frac{1}{x + \sqrt{x^2 + y^2}} \left(1 + \frac{x}{\sqrt{x^2 + y^2}}\right) \].
05

Evaluating at the Given Point

Evaluate \( f_x(x, y) \) at \((3,4)\). First, compute \( \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \). Therefore, \( g(3, 4) = 3 + 5 = 8 \) and \( g_x(3, 4) = 1 + \frac{3}{5} = \frac{8}{5} \). Substitute these into \( f_x(3,4) \): \( f_x(3, 4) = \frac{1}{8} \cdot \frac{8}{5} = \frac{1}{5} \).

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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 powerful tool in calculus used to differentiate composite functions. When dealing with partial derivatives, it helps us differentiate a function that is composed of multiple functions combined together. This is particularly useful when analyzing functions of multiple variables, like in our exercise.
  • Let's break down a composite function: imagine we have a function \( f(x) = h(g(x)) \), where \( g(x) \) is another function inside \( f \).
  • To find the derivative \( f'(x) \) using the chain rule, we first differentiate the outside function \( h \) with respect to \( g \), and then multiply it by the derivative of the inside function \( g \) with respect to \( x \).
  • In the multi-variable context, like \( f(x, y) = \ln(g(x, y)) \), the chain rule is extended to show how \( f \) changes with respect to one variable while keeping others constant.
Utilizing the chain rule simplifies finding the partial derivatives of complex functions, making it easier to handle multi-step differentiation.
natural logarithm
The natural logarithm is a special kind of logarithm featuring the base \( e \), where \( e \approx 2.71828 \). It's often used in calculus due to its simplifying properties when differentiating.
  • The derivative of the natural logarithm \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). This property makes logarithmic differentiation straightforward.
  • In the exercise, the function \( f(x, y) = \ln \left( x + \sqrt{x^2 + y^2} \right) \) requires this derivative property.
  • Logarithms are useful for transforming multiplicative relationships into additive ones, simplifying complex differentiation problems.
Understanding the properties of natural logarithms helps manage their roles in calculus, allowing us to deconstruct functions into manageable parts.
partial differentiation
Partial differentiation is the process of finding the derivative of a function with respect to one variable while keeping other variables constant. It's an essential tool in multivariable calculus.
  • When you have a function like \( f(x, y) \), two partial derivatives exist: with respect to \( x \) and with respect to \( y \).
  • The notation \( f_x(x, y) \) indicates the partial derivative of \( f \) with respect to \( x \).
  • In our step-by-step solution, we calculated \( f_x \) for the function, systematically finding how changes in \( x \) affect \( f(x, y) \) while \( y \) stays constant.
Mastering partial differentiation assists in exploring how variables independently influence functions of multiple variables.
function of two variables
A function of two variables, such as \( f(x, y) \), relates two independent variables \( x \) and \( y \) to a dependent variable \( f \). Such functions describe surfaces in a 3D space where each point is defined by three coordinates.
  • In our exercise, \( f(x, y) = \ln(x + \sqrt{x^2 + y^2}) \) represents how different combinations of \( x \) and \( y \) provide a specific output.
  • These functions can be visualized as surfaces, enabling us to see how changes in \( x \) or \( y \) affect the output.
  • Evaluating partial derivatives like \( f_x \) and \( f_y \) tells us the slope of this surface in the respective directions.
Understanding functions of two variables enriches our ability to model and solve real-world problems involving multiple changing conditions.

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Most popular questions from this chapter

\(45-48\) Assume that all the given functions are differentiable. If \(z=f(x, y),\) where \(x=s+t\) and \(y=s-t,\) show that $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{\partial z}{\partial s} \frac{\partial z}{\partial t}$$

Use a computer to graph the function using various domains and viewpoints. Get a printout that gives a good view of the "peaks and valleys." Would you say the function has a maximum value? Can you identify any points on the graph that you might consider to be "local maximum points"? What about "local minimum points"? $$f(x, y)=3 x-x^{4}-4 y^{2}-10 x y$$

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) \(f(x, y)=1+4 x-5 y, \quad D\) is the closed triangular region with vertices \((0,0),(2,0),\) and \((0,3)\)

(a) How many \(n\) th-order partial derivatives does a function of two variables have? (b) If these partial derivatives are all continuous, how many of them can be distinct? (c) Answer the question in part (a) for a function of three variables.

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=y^{2}-2 y \cos x, \quad 1 \leqslant x \leqslant 7$$
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