Chapter 7: Problem 14
Find the total differential of each function. $$ w=3 x^{2}-x y^{-1}+y^{3} $$
Short Answer
Expert verified
The total differential is \( dw = (6x - \frac{1}{y})dx + (-\frac{x}{y^2} + 3y^2)dy \).
Step by step solution
01
Identify the Function and Variables
First, identify the function given: \( w = 3x^2 - xy^{-1} + y^3 \). The variables in the function are \( x \) and \( y \).
02
Partial Derivative with respect to x
Find the partial derivative of the function \( w \) with respect to \( x \). The derivative is: \( \frac{\partial w}{\partial x} = \frac{d}{dx}(3x^2) - \frac{d}{dx}(xy^{-1}) + \frac{d}{dx}(y^3) = 6x - y^{-1} + 0 = 6x - \frac{1}{y} \).
03
Partial Derivative with respect to y
Find the partial derivative of the function \( w \) with respect to \( y \). The derivative is: \( \frac{\partial w}{\partial y} = \frac{d}{dy}(3x^2) - \frac{d}{dy}(xy^{-1}) + \frac{d}{dy}(y^3) = 0 + x(y^{-2}) + 3y^2 = -\frac{x}{y^2} + 3y^2 \).
04
Write the Total Differential Formula
The total differential of the function \( w \) is given by combining the partial derivatives: \( dw = \frac{\partial w}{\partial x}dx + \frac{\partial w}{\partial y}dy \).
05
Substitute Partial Derivatives into Total Differential
Substitute the partial derivatives found in Steps 2 and 3 into the total differential: \( dw = (6x - \frac{1}{y})dx + (-\frac{x}{y^2} + 3y^2)dy \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are used to find the rate of change of a function with respect to one variable at a time, while keeping all other variables constant. In multivariable functions, knowing how a change in a single variable affects the entire function can be really important. For example.- In the original exercise, to find the partial derivative of the function \( w = 3x^2 - xy^{-1} + y^3 \) with respect to \( x \), we treat \( y \) as a constant.- Similarly, while finding the partial derivative with respect to \( y \), \( x \) is treated as a constant.This helps us understand how changes in \( x \) or \( y \) individually affect the function \( w \). Partial derivatives are denoted by the symbol \( \frac{\partial}{\partial x} \) or \( \frac{\partial}{\partial y} \), indicating that differentiation is occurring with respect to \( x \) or \( y \). The computed partial derivatives then become pieces of the total differential, showing the overall rate of change.
Differentiation
Differentiation is a core concept in calculus that involves calculating the derivative of a function. In the context of multivariable calculus, differentiation extends to include partial derivatives. To perform differentiation:- Identify the function and the variable with respect to which you want to differentiate.- Apply differentiation rules, such as the power rule or the product rule, as needed. In our exercise, differentiation was applied to find the partial derivatives \( \frac{\partial w}{\partial x} = 6x - \frac{1}{y} \) and \( \frac{\partial w}{\partial y} = -\frac{x}{y^2} + 3y^2 \). These derivatives show the rate of change of the function \( w \) with respect to \( x \) and \( y \) respectively. Differentiation is the building block for finding insights about the behavior of a function in response to small changes in variables.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions with more than one variable. Instead of dealing with single-variable functions, we deal with functions such as \( w = 3x^2 - xy^{-1} + y^3 \), which have multiple inputs. Key concepts include:
- Partial derivatives, to examine how each variable independently affects the function.
- Total differentials, providing a way to approximate changes in the function value, encompassing changes in all variables involved.