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find the differential of the function. $$f(x, y)=\sin (x y)$$

Short Answer

Expert verified
The differential is \( dz = y \cos(xy) \, dx + x \cos(xy) \, dy \).

Step by step solution

01

Understand the Problem

We need to find the differential of the function given by \( f(x, y) = \sin(xy) \). The differential gives us a way to approximate changes in the function \( f \) for small changes in \( x \) and \( y \).
02

Recall the Formula for Differential

The differential of a function \( z = f(x, y) \) is given by \( dz = f_x \, dx + f_y \, dy \), where \( f_x \) and \( f_y \) are partial derivatives of \( f \) with respect to \( x \) and \( y \) respectively.
03

Compute the Partial Derivative with Respect to \( x \)

To find \( f_x \), treat \( y \) as a constant and differentiate \( f(x, y) = \sin(xy) \) with respect to \( x \). \[ f_x = \frac{\partial}{\partial x} (\sin(xy)) = y \cos(xy) \]
04

Compute the Partial Derivative with Respect to \( y \)

To find \( f_y \), treat \( x \) as a constant and differentiate \( f(x, y) = \sin(xy) \) with respect to \( y \). \[ f_y = \frac{\partial}{\partial y} (\sin(xy)) = x \cos(xy) \]
05

Write the Differential

Using the partial derivatives found, write the differential as:\[ dz = f_x \, dx + f_y \, dy = y \cos(xy) \, dx + x \cos(xy) \, dy \]

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Key Concepts

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

Partial Derivatives
In multivariable calculus, the concept of partial derivatives extends the idea of ordinary derivatives to functions with two or more variables. This allows us to understand how a function changes while holding all but one of its variables constant. Imagine having a surface representing a mountain range; partial derivatives can tell you how steep the slope is in various directions.For the function given, \( f(x, y) = \sin(xy) \), we can find the partial derivative with respect to \( x \) by treating \( y \) as constant. This gives us how \( f \) changes as \( x \) changes:
  • \( f_x = \frac{\partial}{\partial x}(\sin(xy)) = y \cos(xy) \)
Similarly, the partial derivative with respect to \( y \) treats \( x \) as constant. This yields:
  • \( f_y = \frac{\partial}{\partial y}(\sin(xy)) = x \cos(xy) \)
These derivatives are key to finding the differential, as they tell us how each variable uniquely contributes to changes in the function.
Multivariable Functions
A multivariable function is a type of mathematical function with more than one input variable. Functions of this kind take the form \( f(x, y) \) or \( f(x, y, z) \), allowing us to model complex systems in calculus. For example, temperature over a geographic region can depend on both latitude and longitude.For our function \( f(x, y) = \sin(xy) \), it takes two input variables: \( x \) and \( y \). The output of the function depends on the product of the two variables, which are then inputted into the sine function. This incorporation of multiple variables enables us to study an array of changes in one simultaneously. Multivariable functions are used in fields like physics and engineering for optimization and modeling scenarios where many factors are at play.
Differentials
Differentials in calculus offer a method to estimate how much a function's output changes as its inputs undergo slight changes. Given the function \( f(x, y) = \sin(xy) \), the differential \( dz \) can be calculated using the partial derivatives.The formula for the differential of a function \( z = f(x, y) \) is:
  • \( dz = f_x \, dx + f_y \, dy \)
This formula helps in predicting how a small change in \( x \) and a small change in \( y \) would affect \( z \). By substituting the partial derivatives we found:
  • \( dz = y \cos(xy) \, dx + x \cos(xy) \, dy \)
Thus, differentials provide us a linear approximation of the change around a given point \( (x, y) \), making them a powerful tool in calculus for analyzing functions.

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Most popular questions from this chapter

The tastiness, \(T\), of a soup depends on the volume, \(V\) of the soup in the pot and the quantity, \(S,\) of salt in the soup. If you have more soup, you need more salt to make it taste good. Match the three stories (a)-(c) to the three statements (I)-(III) about partial derivatives. (a) I started adding salt to the soup in the pot. At first the taste improved, but eventually the soup became too salty and continuing to add more salt made it worse. (b) The soup was too salty, so I started adding unsalted soup. This improved the taste at first, but eventually there was too much soup for the salt, and continuing to add unsalted soup just made it worse. (c) The soup was too salty, so adding more salt would have made it taste worse. I added a quart of unsalted soup instead. Now it is not salty enough, but I can improve the taste by adding salt. (I) \(\partial^{2} T / \partial V^{2} < 0\) (II) \(\partial^{2} T / \partial S^{2} < 0\) (III) \(\partial^{2} T / \partial V \partial S > 0\)

Find the gradient at the point. $$f(x, y)=1 /\left(x^{2}+y^{2}\right), \text { at }(-1,3)$$

Are the statements true or false? Give measons for your answer. If \(f(x, y)\) is a constant function, then \(d f=0\)

Give an example of: A function \(f(x, y)\) such that \(f_{x x} \neq 0, f_{y y} \neq 0,\) and \(f_{x y}=0\)

You are climbing a mountain by the steepest route at a slope of \(20^{\circ}\) when you come upon a trail branching off at a \(30^{\circ}\) angle from yours. What is the angle of ascent of the branch trail?

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