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Magazine Chapter 2: Problem 8
Calculate the differential \(d F\) for the given function \(F\). $$ F(x, y)=\tan ^{-1}(x / y)+y^{4} $$
Short Answer
Expert verified
\( dF = \frac{y \, dx - x \, dy}{x^2 + y^2} + 4y^3 \, dy \)
Step by step solution
01
Identify the Components of the Function
The given function is \( F(x, y) = \tan^{-1}(x/y) + y^4 \). We need to find the differential \( dF \), which involves computing the partial derivatives of \( F \) with respect to \( x \) and \( y \).
02
Partial Derivative of F with respect to x
The first term to compute is \( \frac{\partial}{\partial x} \tan^{-1}(x/y) \). By the chain rule, we have \( \frac{1}{1+(x/y)^2} \cdot \frac{1}{y} \), which simplifies to: \( \frac{y}{y^2 + x^2} \). The partial derivative of \( y^4 \) with respect to \( x \) is \( 0 \). Hence, \( \frac{\partial F}{\partial x} = \frac{y}{x^2 + y^2} \).
03
Partial Derivative of F with respect to y
Next, calculate \( \frac{\partial}{\partial y} \tan^{-1}(x/y) \). By the quotient rule and chain rule, it is \( \frac{-x}{x^2 + y^2} \). Additionally, \( \frac{\partial}{\partial y}(y^4) = 4y^3 \). Thus, \( \frac{\partial F}{\partial y} = \frac{-x}{x^2 + y^2} + 4y^3 \).
04
Write the Expression for the Differential dF
The differential \( dF \) is given by \( dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy \). Substituting the partial derivatives, we get: \[ dF = \frac{y}{x^2 + y^2}dx + \left(\frac{-x}{x^2 + y^2} + 4y^3\right)dy \].
05
Simplify the Expression
Upon combining and simplifying, the differential remains as derived: \[ dF = \frac{y \, dx - x \, dy}{x^2 + y^2} + 4y^3 \, dy \]. This is the differential of the given function \( F(x, y) \).
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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 a fundamental concept in multivariable calculus. They focus on how a function changes as one of the variables is varied, while others are held constant.
For instance, when dealing with a function like \( F(x, y) = \tan^{-1}(x/y) + y^4 \), we find the partial derivatives \( \frac{\partial F}{\partial x} \) and \( \frac{\partial F}{\partial y} \).
For instance, when dealing with a function like \( F(x, y) = \tan^{-1}(x/y) + y^4 \), we find the partial derivatives \( \frac{\partial F}{\partial x} \) and \( \frac{\partial F}{\partial y} \).
- \( \frac{\partial F}{\partial x} \): examines changes in \( F \) as \( x \) changes, with \( y \) constant.
- \( \frac{\partial F}{\partial y} \): examines changes in \( F \) as \( y \) changes, with \( x \) constant.
Chain Rule
The chain rule is a key principle for differentiating composite functions. It helps break down derivatives into parts that are easier to handle.
In the expression \( \tan^{-1}(x/y) \), the function is composed of two parts: the arctangent function and the fraction \( x/y \). The chain rule allows us to consider how changes in \( x \) and \( y \) impact \( \tan^{-1}(x/y) \).
We first differentiate \( \tan^{-1}(u) \) with respect to \( u \), and then the fraction with respect to \( x \) and \( y \). This step is crucial in the partial derivative calculations, as seen in \( \frac{y}{x^2 + y^2} \) for \( x \), and \( \frac{-x}{x^2 + y^2} \) for \( y \).
In the expression \( \tan^{-1}(x/y) \), the function is composed of two parts: the arctangent function and the fraction \( x/y \). The chain rule allows us to consider how changes in \( x \) and \( y \) impact \( \tan^{-1}(x/y) \).
We first differentiate \( \tan^{-1}(u) \) with respect to \( u \), and then the fraction with respect to \( x \) and \( y \). This step is crucial in the partial derivative calculations, as seen in \( \frac{y}{x^2 + y^2} \) for \( x \), and \( \frac{-x}{x^2 + y^2} \) for \( y \).
Quotient Rule
The quotient rule complements the chain rule for functions that appear as a ratio of two differentiable functions. It is given by:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]Using it helps differentiate expressions like \( x/y \).
When calculating \( \frac{\partial}{\partial y} \tan^{-1}(x/y) \), we use this rule to differentiate the fraction, combining it with the result from differentiating \( \tan^{-1} \).
When calculating \( \frac{\partial}{\partial y} \tan^{-1}(x/y) \), we use this rule to differentiate the fraction, combining it with the result from differentiating \( \tan^{-1} \).
- This results in terms like \( \frac{-x}{x^2 + y^2} \), crucial in constructing \( dF \).
Differentials
Differentials extend the concept of derivatives. They demonstrate how a small change in input impacts the output of a function.
The differential \( dF \) of a function \( F(x, y) \) is formally expressed as:\[ dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy \]Here, \( dx \) and \( dy \) represent small changes in \( x \) and \( y \).
With \( F(x, y) = \tan^{-1}(x/y) + y^4 \), after calculating partial derivatives, we find:
The differential \( dF \) of a function \( F(x, y) \) is formally expressed as:\[ dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy \]Here, \( dx \) and \( dy \) represent small changes in \( x \) and \( y \).
With \( F(x, y) = \tan^{-1}(x/y) + y^4 \), after calculating partial derivatives, we find:
- \( dF = \frac{y}{x^2 + y^2}dx + \left(\frac{-x}{x^2 + y^2} + 4y^3\right)dy \)
