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Chapter 16: Problem 4

Find the gradient of \(f\) at \(P\). \(f(x, y)=x \ln (x-y): \quad P(5,4)\)

Short Answer

Expert verified
The gradient at \( P(5,4) \) is \( (5, -5) \).

Step by step solution

01

Identify the Function and Point

The function given is \( f(x, y) = x \ln(x-y) \). We need to find the gradient of this function at the point \( P(5, 4) \).
02

Calculate Partial Derivative with Respect to x

Differentiate the function \( f(x, y) = x \ln(x-y) \) with respect to \( x \). Using the product rule, we have: \[ \frac{\partial f}{\partial x} = \ln(x-y) + \frac{x}{x-y}\cdot 1 = \ln(x-y) + \frac{x}{x-y} \]
03

Calculate Partial Derivative with Respect to y

Differentiate the function \( f(x, y) = x \ln(x-y) \) with respect to \( y \). This gives: \[ \frac{\partial f}{\partial y} = x\cdot \frac{1}{x-y} \cdot (-1) = -\frac{x}{x-y} \]
04

Evaluate Partial Derivatives at P

Substitute \( x = 5 \) and \( y = 4 \) into the partial derivatives.1. \( \frac{\partial f}{\partial x} = \ln(5-4) + \frac{5}{5-4} = \ln(1) + 5 = 0 + 5 = 5 \)2. \( \frac{\partial f}{\partial y} = -\frac{5}{5-4} = -5 \)
05

Write the Gradient Vector

The gradient of \( f \) at point \( P(5, 4) \) is:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (5, -5) \]

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Key Concepts

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

Partial Derivative
A partial derivative is a technique used in calculus to find the rate of change of a function with respect to one of its variables, while keeping all other variables constant. This is useful when dealing with functions of multiple variables, like our function \( f(x, y) = x \ln(x-y) \). To calculate a partial derivative:
  • Focus on one variable at a time. Treat all other variables as constants.
  • Differentiation is performed similarly to single-variable calculus.
In the solution, we started with the partial derivative of \( f \) with respect to \( x \), leading to the expression \( \ln(x-y) + \frac{x}{x-y} \). Similarly, the partial derivative with respect to \( y \) gave us \( -\frac{x}{x-y} \). These derivatives helped build towards finding the gradient.
Gradient Vector
The gradient vector is a vital tool in multivariable calculus. It consists of all the partial derivatives of a function. In simple terms, it points in the direction of the steepest increase of the function. For a function \( f(x, y) \), the gradient is represented as:
  • \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]
  • It's a vector showing how \( f \) changes at a point, directionally.
In our exercise, after calculating the partial derivatives, we compiled them into the gradient vector at point \( P(5, 4) \), which is \( (5, -5) \). This tells us that at \( P \), the function increases in the \( x \)-direction and decreases in the \( y \)-direction.
Product Rule
The product rule is a differentiation rule used when finding the derivative of a product of two functions. If you have functions \( u(x) \) and \( v(x) \), then:
  • The derivative of their product \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \).
For our function \( f(x, y) = x \ln(x-y) \), \( x \) is \( u(x) \) and \( \ln(x-y) \) is \( v(x) \). Thus, applying the product rule results in the partial derivative \( \ln(x-y) + \frac{x}{x-y} \) when differentiating with respect to \( x \). It helps us understand each term's individual contribution when differentiating complex expressions.
Evaluation at a Point
After computing derivatives or other function-related information, substituting specific values can be crucial. It helps in finding specific characteristics, like the gradient at a particular point. In this exercise, we assessed the partial derivatives at \( P(5, 4) \). This step involves:
  • Substitute the given \( x \) and \( y \) values into the derived equations.
  • Calculate the numerical value that represents change at that point.
For \( \frac{\partial f}{\partial x} \), input \( x = 5 \) and \( y = 4 \) to yield \( 5 \). For \( \frac{\partial f}{\partial y} \), it results in \( -5 \). These specific evaluations ultimately formed the gradient vector \( (5, -5) \) at \( P \). By evaluating at a point, we acquire precise information about the function's behavior at that location.

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Most popular questions from this chapter

Use Lagrange multipliers to find the extrema of \(f\) subject to the stated constraints. $$ \begin{aligned} f(x, y, z)=4 x^{2}+y^{2}+z^{2} & \\ 2 x-y+z=4, & x+2 y-z=1 \end{aligned} $$

Use the method of least squares (see Exercise 45 ) to find a line \(y=m x+b\) that best fits the given data \(\begin{array}{lllll}x \text { values } & 1 & 4 & 6 & 8 \\ y \text { values } & 1 & 3 & 2 & 4\end{array}\)

Use the following formulas with \(h=0.01\) to approximate \(f_{x x}(0.6,0.8)\) and \(f_{x y}(0.6,0.8)\) $$ \begin{array}{l} f_{x x}(x, y) \approx \frac{f(x+h, y)-2 f(x, y)+f(x-h, y)}{h^{2}} \\ f_{y y}(x, y) \approx \frac{f(x, y+h)-2 f(x, y)+f(x, y-h)}{h^{2}} \end{array} $$ \(f(x, y)=\sec ^{2}\left[\tan \left(x y^{2}\right)\right] \sin (x y)\)

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Find the first partial derivatives of \(f\). $$ f(r, s, v, p)=r^{3} \tan s+\sqrt{s e^{\left(v^{2}\right)}}-v \cos 2 p $$
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