Problem 30
Question
\(25-30\) Find the differential of the function. $$ w=x y e^{x z} $$
Step-by-Step Solution
Verified Answer
The differential is \(dw = \left(y e^{x z} + xyz e^{x z}\right) \, dx + \left(x e^{x z}\right) \, dy + \left(x^2y e^{x z}\right) \, dz\).
1Step 1: Understand the problem
Find the differential \(dw\) for the function \(w = x y e^{x z}\). The differential of a function of multiple variables can be obtained by taking the partial derivatives with respect to each variable in turn.
2Step 2: Partial derivative with respect to x
Take the partial derivative of \(w\) with respect to \(x\). Apply the product rule and chain rule:\[\frac{\partial w}{\partial x} = y e^{x z} + xyz e^{x z}\]First, differentiate \(x\), holding \(y\) and \(e^{xz}\) constant, then differentiate \(e^{xz}\) with respect to \(x\), treating \(y\) as constant.
3Step 3: Partial derivative with respect to y
Take the partial derivative of \(w\) with respect to \(y\):\[\frac{\partial w}{\partial y} = x e^{x z}\]Here, differentiate with respect to \(y\), treating \(x\) and \(e^{xz}\) as constants.
4Step 4: Partial derivative with respect to z
Take the partial derivative of \(w\) with respect to \(z\):\[\frac{\partial w}{\partial z} = xyxe^{x z}\]Differentiate \(e^{xz}\) with respect to \(z\) while treating \(xy\) as constant.
5Step 5: Compute the differential dw
Combine all partial derivatives to find the differential \(dw\):\[dw = \left(y e^{x z} + xyz e^{x z}\right) \, dx + \left(x e^{x z}\right) \, dy + \left(x^2y e^{x z}\right) \, dz\]Each term corresponds to the partial derivative times the differential of the respective variable.
Key Concepts
Partial DerivativesProduct RuleChain RuleMultivariable Calculus
Partial Derivatives
In multivariable calculus, partial derivatives are an essential concept when dealing with functions of more than one variable. They represent how a function changes as just one of the variables changes, while keeping the other variables constant. For example, if you have a function \(w = x y e^{x z}\), you want to know how \(w\) changes as \(x\), \(y\), or \(z\) changes independently.
- To find a partial derivative, you calculate the derivative of the function with respect to one variable while treating the others as constants.
- It's like looking at one slice of the function at a time, seeing how that slice bends or varies.
In the given function, \(w = x y e^{x z}\), we found three partial derivatives, one for each variable \(x\), \(y\), and \(z\). Each of these gives us a look at how the function changes as that particular variable changes, providing crucial insights for calculating the differential.
- To find a partial derivative, you calculate the derivative of the function with respect to one variable while treating the others as constants.
- It's like looking at one slice of the function at a time, seeing how that slice bends or varies.
In the given function, \(w = x y e^{x z}\), we found three partial derivatives, one for each variable \(x\), \(y\), and \(z\). Each of these gives us a look at how the function changes as that particular variable changes, providing crucial insights for calculating the differential.
Product Rule
The product rule is a fundamental tool when differentiating functions that are composed of the product of two or more functions. It's frequently used in calculus when dealing with partial derivatives a product of functions. Consider a scenario where your function is the product of two functions, \(u(x)\) and \(v(x)\). The product rule states:
\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]
In our example, \(w = x y e^{x z}\), to find the partial derivative with respect to \(x\), we apply the product rule multiple times. When differentiating \(x y e^{x z}\), we consider \(x\) and \(y e^{x z}\), treating \(y\) and \(e^{x z}\) as constants at first, and then apply the product rule again when differentiating \(e^{x z}\) with respect to \(x\).
This results in the partial derivative formula \(\frac{\partial w}{\partial x} = y e^{x z} + xyz e^{x z}\), showcasing how the product rule plays a vital role in handling more complex derivatives.
\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]
In our example, \(w = x y e^{x z}\), to find the partial derivative with respect to \(x\), we apply the product rule multiple times. When differentiating \(x y e^{x z}\), we consider \(x\) and \(y e^{x z}\), treating \(y\) and \(e^{x z}\) as constants at first, and then apply the product rule again when differentiating \(e^{x z}\) with respect to \(x\).
This results in the partial derivative formula \(\frac{\partial w}{\partial x} = y e^{x z} + xyz e^{x z}\), showcasing how the product rule plays a vital role in handling more complex derivatives.
Chain Rule
Another crucial rule in calculus is the chain rule, which helps us differentiate composite functions. A composite function is when one function is inside of another, such as \(f(g(x))\). The chain rule states:
\[\frac{d}{dx}f(g(x)) = f'(g(x)) g'(x)\]
In our example function \(w = x y e^{x z}\), the term \(e^{x z}\) is a perfect example of a composite function. To find the derivative of this function, we treat \(e^\cdot\) as the outer function \(f\) and \(x z\) as the inner function \(g\). The chain rule allows us to differentiate \(e^{x z}\) by first differentiating \(e^u\) where \(u = x z\), which gives \(e^u\), and then multiplying by the derivative of \(u\) with respect to \(x\), which is \(z\).
Thus, in the step where we found \(\frac{\partial w}{\partial x}\), the chain rule was used to handle the differentiation of \(e^{x z}\). This rule plays a key role in breaking down complex dependent functions into manageable parts.
\[\frac{d}{dx}f(g(x)) = f'(g(x)) g'(x)\]
In our example function \(w = x y e^{x z}\), the term \(e^{x z}\) is a perfect example of a composite function. To find the derivative of this function, we treat \(e^\cdot\) as the outer function \(f\) and \(x z\) as the inner function \(g\). The chain rule allows us to differentiate \(e^{x z}\) by first differentiating \(e^u\) where \(u = x z\), which gives \(e^u\), and then multiplying by the derivative of \(u\) with respect to \(x\), which is \(z\).
Thus, in the step where we found \(\frac{\partial w}{\partial x}\), the chain rule was used to handle the differentiation of \(e^{x z}\). This rule plays a key role in breaking down complex dependent functions into manageable parts.
Multivariable Calculus
Multivariable calculus is the extension of calculus to functions of more than one variable. Unlike with single-variable calculus, you're often dealing with surfaces or higher-dimensional spaces rather than simple curves.
- The concept of differentials, as seen in the provided exercise, helps understand how small changes in input variables affect the output of a multivariable function.
- It uses partial derivatives to analyze the rate of change along each variable direction.
For the function \(w = x y e^{x z}\), we leveraged partial derivatives with respect to each variable to understand how changes in \(x\), \(y\), and \(z\) individually affect \(w\).
In multivariable calculus, calculating the differential \(dw = \left(y e^{x z} + xyz e^{x z}\right)\, dx + \left(x e^{x z}\right)\, dy + \left(x^2 y e^{x z}\right)\, dz\) demonstrates summarizing these small changes into a single expression. This differential combines all angles of change, offering a thorough understanding of how \(w\) responds to variations in its variables. Multivariable calculus equips us with the tools to handle and interpret these complex interactions.
- The concept of differentials, as seen in the provided exercise, helps understand how small changes in input variables affect the output of a multivariable function.
- It uses partial derivatives to analyze the rate of change along each variable direction.
For the function \(w = x y e^{x z}\), we leveraged partial derivatives with respect to each variable to understand how changes in \(x\), \(y\), and \(z\) individually affect \(w\).
In multivariable calculus, calculating the differential \(dw = \left(y e^{x z} + xyz e^{x z}\right)\, dx + \left(x e^{x z}\right)\, dy + \left(x^2 y e^{x z}\right)\, dz\) demonstrates summarizing these small changes into a single expression. This differential combines all angles of change, offering a thorough understanding of how \(w\) responds to variations in its variables. Multivariable calculus equips us with the tools to handle and interpret these complex interactions.
Other exercises in this chapter
Problem 30
Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$\begin{array}{l}{f(x, y)=3+x y-x-2 y, \quad D \text { is the closed triangular }} \\
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\(29-38\) Determine the set of points at which the function is continuous. $$F(x, y)=\frac{x-y}{1+x^{2}+y^{2}}$$
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Find the first partial derivatives of the function. $$f(x, y, z)=x \sin (y-z)$$
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Near a buoy, the depth of a lake at the point with coordinates \((x, y)\) is \(z=200+0.02 x^{2}-0.001 y^{3},\) where \(x, y,\) and \(z\) are measured in meters.
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