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Mobile App Chapter 4: Problem 17
Bilden Sie die partiellen Ableitwngen 1. Ordnung der jeweiligen Funktion \(z=f(x ; y)\) mit \(x=x(u ; v), y=y(u ; v)\) nach den Parametern \(u\) bzw, \(v\) unter Verwendung der Kettenregel: a) \(z=\tan (x+y)\) mit \(x=u^{2}+v, y=u^{2}-v\) b) \(\quad z=x^{2} y^{3}+x^{3} y^{2} \quad\) mit \(x=u+v, y=u-v\)
Short Answer
Expert verified
Part a: \(\frac{\partial z}{\partial u} = 4u \sec^2(x + y)\), \(\frac{\partial z}{\partial v} = 0\). Part b: Apply the chain rule with derived formulas for \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) inserting \(x = u + v\) and \(y = u - v\).
Step by step solution
01
Understanding the Chain Rule for Partial Derivatives
The problem requires finding the first-order partial derivatives of the function using the chain rule. The chain rule for functions of multiple variables involves taking the derivative with respect to each parameter and considering the dependencies.
02
Compute Partial Derivatives of z with respect to x and y for part a
Given function: \[z = \tan(x + y)\]Compute \(\frac{\partial z}{\partial x}\):\[\frac{\partial z}{\partial x} = \sec^2(x + y)\]Compute \(\frac{\partial z}{\partial y}\):\[\frac{\partial z}{\partial y} = \sec^2(x + y)\]
03
Apply the Chain Rule for part a
Use chain rule formulas given the dependencies \(x = u^2 + v\) and \(y = u^2 - v\)Compute \(\frac{\partial z}{\partial u}\):\[\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}\]i.e.,\[\frac{\partial z}{\partial u} = \sec^2(x + y) (2u + 2u) = 4u \sec^2(x + y)\]Compute \(\frac{\partial z}{\partial v}\):\[\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}\]i.e.,\[\frac{\partial z}{\partial v} = \sec^2(x + y) (1 + (-1)) = 0\]
04
Compute Partial Derivatives of z with respect to x and y for part b
Given function: \[z = x^2 y^3 + x^3 y^2\]Compute \(\frac{\partial z}{\partial x}\):\[\frac{\partial z}{\partial x} = 2xy^3 + 3x^2 y^2\]Compute \(\frac{\partial z}{\partial y}\):\[\frac{\partial z}{\partial y} = 3x^2 y^2 + 2x^3 y\]
05
Apply the Chain Rule for part b
Use chain rule formulas given the dependencies \(x = u + v\) and \(y = u - v\)Compute \(\frac{\partial z}{\partial u}\):\[\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}\]i.e.,\[\frac{\partial z}{\partial u} = (2x y^3 + 3 x^2 y^2)(1) + (3 x^2 y^2 + 2x^3 y)(1)\]Plug in \(x = u + v\) and \(y = u - v\):\[\frac{\partial z}{\partial u} = (2(u + v)(u - v)^3 + 3(u + v)^2(u - v)^2) + (3(u + v)^2(u - v)^2 + 2(u + v)^3(u - v))\]Compute \(\frac{\partial z}{\partial v}\):\[\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}\]i.e.,\[\frac{\partial z}{\partial v} = (2x y^3 + 3 x^2 y^2)(1) + (3 x^2 y^2 + 2x^3 y)(-1)\]Plug in \(x = u + v\) and \(y = u - v\):\[\frac{\partial z}{\partial v} = (2(u + v)(u - v)^3 + 3(u + v)^2(u - v)^2) - (3(u + v)^2(u - v)^2 + 2(u + v)^3(u - v))\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
chain rule in calculus
The chain rule is a formula for computing the derivative of the composition of two or more functions. When dealing with multivariable calculus, the chain rule helps us differentiate a function with respect to one variable while considering its dependency on other variables. Imagine a function \(z\) depends on \(x\) and \(y\), and \(x\) and \(y\) depend on \(u\) and \(v\). To find \(\frac{\partial z}{\partial u}\), you consider how \(z\) changes with respect to \(x\) and \(y\), and then how \(x\) and \(y\) change with respect to \(u\). This leads us to use the formula:
\( \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u} \).
\( \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u} \).
- Use the partial derivative of \(z\) with respect to \(x\), \( \frac{\partial z}{\partial x} \).
- Multiply it by the partial derivative of \(x\) with respect to \(u\), \( \frac{\partial x}{\partial u} \).
- Add the product to the partial derivative of \(z\) with respect to \(y\), \( \frac{\partial z}{\partial y} \), multiplied by the partial derivative of \(y\) with respect to \(u\), \( \frac{\partial y}{\partial u} \).
partial derivatives
Partial derivatives are used to measure how a function changes as only one of the variables is varied. For a function \( z = f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial z}{\partial x} \). This represents how \( z \) changes when \( x \) changes, keeping \( y \) constant.
Partial derivatives are crucial in multivariable calculus as functions often depend on more than one variable. Here’s an example:
For \( z = x^2 y^3 + x^3 y^2 \), to find \( \frac{\partial z}{\partial x} \), we differentiate \( z \) with respect to \( x \) while treating \( y \) as a constant:
\( \frac{\partial z}{\partial x} = 2x y^3 + 3x^2 y^2 \).
Similarly, to find \( \frac{\partial z}{\partial y} \):
\( \frac{\partial z}{\partial y} = 3x^2 y^2 + 2x^3 y \).
Partial derivatives are crucial in multivariable calculus as functions often depend on more than one variable. Here’s an example:
For \( z = x^2 y^3 + x^3 y^2 \), to find \( \frac{\partial z}{\partial x} \), we differentiate \( z \) with respect to \( x \) while treating \( y \) as a constant:
\( \frac{\partial z}{\partial x} = 2x y^3 + 3x^2 y^2 \).
Similarly, to find \( \frac{\partial z}{\partial y} \):
\( \frac{\partial z}{\partial y} = 3x^2 y^2 + 2x^3 y \).
multivariable calculus
Multivariable calculus extends calculus to functions of several variables. In this context, we study functions of the form \( f(x, y, z, ...) \). The goal is to understand how these functions change as any of the variables change. To do this, we use partial derivatives, gradient vectors, and other tools.
When working with functions of several variables, you need to consider:
When working with functions of several variables, you need to consider:
- Partial derivatives: Differentiate the function with respect to one variable at a time.
- Gradient: A vector that consists of all the partial derivatives of a function. It shows the direction of the steepest ascent.
- Chain rule: Helps determine how changes in independent variables (e.g., \(u, v\)) affect dependent variables (e.g., \(x, y\)).
functions of multiple variables
Functions of multiple variables depend on two or more inputs. For example, a function can be represented as \( z = f(x, y) \), where \( z \) depends on both \( x \) and \( y \). Understanding these functions requires knowing how each variable influences the function's outcome.
Key points to remember include:
Key points to remember include:
- Visualize the function: Graphs, contour plots, and surface plots help visualize how the function behaves with different variable changes.
- Evaluate partial derivatives: Determine the rate of change of the function with respect to each variable independently. Use symbols like \( \frac{\partial z}{\partial x} \) to denote these rates.
- Use the chain rule: To find how changes in parameters affect the function, especially when parameters are interdependent.
