Chapter 13: Problem 110
Find the indicated partial derivatives. $$ \begin{aligned} &V=x e^{2 x-y}+w e^{z w}+y w \\ &\partial V / \partial x, \partial V / \partial y, \partial V / \partial z, \partial V / \partial w \end{aligned} $$
Short Answer
Expert verified
\( \frac{\partial V}{\partial x} = e^{2x-y}(1+2x) \), \( \frac{\partial V}{\partial y} = -x e^{2x-y} + w \), \( \frac{\partial V}{\partial z} = w^2 e^{zw} \), \( \frac{\partial V}{\partial w} = e^{zw} + zw e^{zw} + y \).
Step by step solution
01
Understanding the Problem
We need to find the partial derivatives of the function \( V = x e^{2x-y} + w e^{zw} + yw \) with respect to each of the variables \( x, y, z, \) and \( w \). This means taking the derivative of \( V \) while treating all other variables as constants.
02
Calculate \( \frac{\partial V}{\partial x} \)
Differentiate \( V \) with respect to \( x \):- The term \( x e^{2x-y} \) contributes \( \frac{d}{dx}[x e^{2x-y}] = e^{2x-y} + 2x e^{2x-y} \) using the product rule.- Other terms are constants in terms of \( x \) and contribute 0.Thus, \[ \frac{\partial V}{\partial x} = e^{2x-y} + 2x e^{2x-y} \] which simplifies to \( e^{2x-y} (1 + 2x) \).
03
Calculate \( \frac{\partial V}{\partial y} \)
Differentiate \( V \) with respect to \( y \):- The term \( x e^{2x-y} \) gives \( -x e^{2x-y} \) when differentiated with respect to \( y \), as \( y \) appears as \( -y \) in the exponent.- The term \( yw \) gives \( w \).- Other terms are constants in terms of \( y \) and contribute 0.Thus, \[ \frac{\partial V}{\partial y} = -x e^{2x-y} + w \].
04
Calculate \( \frac{\partial V}{\partial z} \)
Differentiate \( V \) with respect to \( z \):- Only the term \( w e^{zw} \) depends on \( z \), and gives \( w^2 e^{zw} \).- Other terms are constants in terms of \( z \) and contribute 0.Thus, \[ \frac{\partial V}{\partial z} = w^2 e^{zw} \].
05
Calculate \( \frac{\partial V}{\partial w} \)
Differentiate \( V \) with respect to \( w \):- The term \( w e^{zw} \) gives \( e^{zw} + zw e^{zw} \) using the product rule.- The term \( yw \) gives \( y \).- The term \( x e^{2x-y} \) is constant in terms of \( w \) and contributes 0.Thus, \[ \frac{\partial V}{\partial w} = e^{zw} + zw e^{zw} + y \].
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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 a branch of calculus focusing on functions of multiple variables. Unlike single-variable calculus where functions have just one input and one output, multivariable functions can have multiple inputs and outputs. These are often visualized as surfaces or higher-dimensional forms.
In this exercise, we're dealing with a function \( V \) that depends on four variables: \( x, y, z, \) and \( w \). The function \( V(x, y, z, w) = x e^{2x-y} + w e^{zw} + yw \) presents a scenario where we are interested in how \( V \) changes with respect to each variable independently. This is done by calculating partial derivatives.
In this exercise, we're dealing with a function \( V \) that depends on four variables: \( x, y, z, \) and \( w \). The function \( V(x, y, z, w) = x e^{2x-y} + w e^{zw} + yw \) presents a scenario where we are interested in how \( V \) changes with respect to each variable independently. This is done by calculating partial derivatives.
- **Partial derivatives** take the derivative of a function concerning just one variable, keeping others constant. This is crucial in fields where systems are affected by several simultaneous inputs.
- **Real-world applications** of multivariable calculus include optimizing functions in economics, calculating pressure and temperature changes in physics, and analyzing spatial designs in engineering.
Product Rule
The product rule is a fundamental tool in calculus used when differentiating products of functions. If you have a function \( u(x) \cdot v(x) \), the product rule states that its derivative is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This extends to partial derivatives in multivariable scenarios.
In our problem, when taking the partial derivative \( \frac{\partial V}{\partial x} \) of \( x e^{2x-y} \), we apply the product rule:
In our problem, when taking the partial derivative \( \frac{\partial V}{\partial x} \) of \( x e^{2x-y} \), we apply the product rule:
- **Differentiate \( x \)**: It becomes \( 1 \), while keeping \( e^{2x-y} \) constant.
- **Differentiate \( e^{2x-y} \)**: The derivative is \( e^{2x-y} \cdot 2 \) with respect to \( x \), while keeping \( x \) constant.
- **Combine results**: This gives \( e^{2x-y} + 2x e^{2x-y} \), which simplifies to \( e^{2x-y}(1 + 2x) \).
Exponential Functions
Exponential functions are mathematical functions of the form \( e^{x} \), where \( e \approx 2.718 \) is Euler's number, a fundamental constant in mathematics. These functions have a rate of growth proportional to their current value and are ubiquitous in growth and decay models, finance, and scientific contexts.
In the function \( V(x, y, z, w) = x e^{2x-y} + w e^{zw} + yw \), exponential expressions appear in two places, complicating the differentiation process. Keep in mind:
In the function \( V(x, y, z, w) = x e^{2x-y} + w e^{zw} + yw \), exponential expressions appear in two places, complicating the differentiation process. Keep in mind:
- **The chain rule**: Often accompanies exponential derivatives. If you have \( e^{g(x)} \), its derivative is \( e^{g(x)} g'(x) \).
- **Rapid growth/decay**: This property makes them vital in modeling natural phenomena, such as population growth where resources multiply steadily.
- **Stability**: Despite complex changes in variables, expressions containing \( e \) retain predictable structural properties.