Problem 14
Question
In Exercises 13 through 24 , find the indicated partial derivatives by holding all but one of the variables constant and applying theorems for ordinary differentiation. $$ f(x, y)=\frac{x+y}{\sqrt{y^{2}-x^{2}}} ; D_{2} f(x, y) $$
Step-by-Step Solution
Verified Answer
\( D_{2} f(x, y) = \frac{-x^2 - xy}{(y^2 - x^2)\sqrt{y^2 - x^2}} \)
1Step 1 - Identify the function and the indicated partial derivative
The given function is \( f(x, y) = \frac{x+y}{\sqrt{y^2 - x^2}} \). The exercise asks for \( D_{2} f(x, y) \), which means we need to find the partial derivative of the function with respect to \( y \).
2Step 2 - Apply the quotient rule
To find the partial derivative with respect to \( y \), use the quotient rule: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]. Here, \( u = x + y \) and \( v = \sqrt{y^2 - x^2} \).
3Step 3 - Differentiate the numerator
The numerator \( u = x + y \). Its partial derivative with respect to \( y \) is \( \frac{\partial}{\partial y}(x + y) = 0 + 1 = 1 \).
4Step 4 - Differentiate the denominator
The denominator \( v = \sqrt{y^2 - x^2} \). Its partial derivative with respect to \( y \) involves the chain rule: \( \frac{\partial}{\partial y}(\sqrt{y^2 - x^2}) = \frac{1}{2}(y^2 - x^2)^{-1/2} \cdot 2y = \frac{y}{\sqrt{y^2 - x^2}} \).
5Step 5 - Apply the quotient rule
Using the results from Steps 3 and 4, apply the quotient rule: \[ \frac{\partial}{\partial y} \left( \frac{x+y}{\sqrt{y^2 - x^2}} \right) = \frac{(1)\sqrt{y^2 - x^2} - (x+y)\frac{y}{\sqrt{y^2 - x^2}}}{(\sqrt{y^2 - x^2})^2} \].
6Step 6 - Simplify the expression
Simplify the expression: \[ \frac{\sqrt{y^2 - x^2} - \frac{(x+y)y}{\sqrt{y^2 - x^2}}}{y^2 - x^2} = \frac{(y^2 - x^2) - (x+y)y}{(y^2 - x^2)\sqrt{y^2 - x^2}} \].
7Step 7 - Combine and reduce terms
Combine and reduce the terms in the numerator: \[ (y^2 - x^2) - xy - y^2 = -x^2 - xy \], so the partial derivative is \[ \frac{-x^2 - xy}{(y^2 - x^2)\sqrt{y^2 - x^2}} \].
Key Concepts
Quotient RuleChain RulePartial Differentiation
Quotient Rule
When dealing with functions that are fractions, specifically where one function is divided by another, the quotient rule is an invaluable tool for differentiation. The quotient rule states that for two functions, say \( u \) and \( v \), their derivative can be found using the formula:
\[ \frac{d}{dx}\frac{u}{v} = \frac{u'v - uv'}{v^2} \]
The first step is to identify the numerator (\( u \)) and the denominator (\( v \)). In our given function \( f(x, y) = \frac{x+y}{\text{sqrt}(y^2 - x^2)} \), the numerator is \( u = x + y \) and the denominator is \( v = \text{sqrt}(y^2 - x^2) \).
Once identified, differentiate both parts separately. For \( u \), the partial derivative with respect to \( y \) is \( \frac{\text{du}}{\text{dy}} = 1 \), since \( x \) is treated as a constant. For \( v \), the chain rule will be necessary for differentiation, as described in the next section.
After obtaining the partial derivatives, apply the quotient rule formula by plugging \( u \), \( u' \), \( v \), and \( v' \) into it. The final step involves simplifying the resulting expression to get the partial derivative.
\[ \frac{d}{dx}\frac{u}{v} = \frac{u'v - uv'}{v^2} \]
The first step is to identify the numerator (\( u \)) and the denominator (\( v \)). In our given function \( f(x, y) = \frac{x+y}{\text{sqrt}(y^2 - x^2)} \), the numerator is \( u = x + y \) and the denominator is \( v = \text{sqrt}(y^2 - x^2) \).
Once identified, differentiate both parts separately. For \( u \), the partial derivative with respect to \( y \) is \( \frac{\text{du}}{\text{dy}} = 1 \), since \( x \) is treated as a constant. For \( v \), the chain rule will be necessary for differentiation, as described in the next section.
After obtaining the partial derivatives, apply the quotient rule formula by plugging \( u \), \( u' \), \( v \), and \( v' \) into it. The final step involves simplifying the resulting expression to get the partial derivative.
Chain Rule
The chain rule is essential for differentiating composite functions. In essence, it provides a way to differentiate functions nested within other functions.
For a function \( v(y) = \text{sqrt}(y^2 - x^2) \), differentiating with respect to \( y \) requires the inner function \( y^2 - x^2 \), which is inside the square root. The chain rule instructs us to first differentiate the outer function and then multiply by the derivative of the inner function.
Mathematically, this can be written as:
\[ \frac{\text{d}}{\text{dy}}\text{sqrt}(y^2 - x^2) = \frac{1}{2}(y^2 - x^2)^{-1/2} \times 2y = \frac{y}{\text{sqrt}(y^2 - x^2)} \]
You calculate the derivative of the outer square root function as \( \frac{1}{2}(something)^{-1/2} \), and then stick the inner function \( y^2 - x^2 \) inside. Next, you multiply by the derivative of \( y^2 - x^2 \), which is \( 2y \). Simplifying, you get \( \frac{y}{\text{sqrt}(y^2 - x^2)} \), which can then be used in the quotient rule's formula.
For a function \( v(y) = \text{sqrt}(y^2 - x^2) \), differentiating with respect to \( y \) requires the inner function \( y^2 - x^2 \), which is inside the square root. The chain rule instructs us to first differentiate the outer function and then multiply by the derivative of the inner function.
Mathematically, this can be written as:
\[ \frac{\text{d}}{\text{dy}}\text{sqrt}(y^2 - x^2) = \frac{1}{2}(y^2 - x^2)^{-1/2} \times 2y = \frac{y}{\text{sqrt}(y^2 - x^2)} \]
You calculate the derivative of the outer square root function as \( \frac{1}{2}(something)^{-1/2} \), and then stick the inner function \( y^2 - x^2 \) inside. Next, you multiply by the derivative of \( y^2 - x^2 \), which is \( 2y \). Simplifying, you get \( \frac{y}{\text{sqrt}(y^2 - x^2)} \), which can then be used in the quotient rule's formula.
Partial Differentiation
In partial differentiation, we differentiate a multivariable function with respect to one variable while treating all other variables as constants. The notation typically used is \( D_2 f(x, y) \) for the partial derivative with respect to \( y \).
For our function \( f(x, y) = \frac{x + y}{\text{sqrt}(y^2 - x^2)} \), finding \( D_2 f(x, y) \) means differentiating \( f \) with respect to \( y \).
The steps for partial differentiation here include:
\[ \frac{-x^2 - xy}{(y^2 - x^2)\text{sqrt}(y^2 - x^2)} \].Understanding these techniques build a foundation for tackling complex multivariable calculus problems effectively.
For our function \( f(x, y) = \frac{x + y}{\text{sqrt}(y^2 - x^2)} \), finding \( D_2 f(x, y) \) means differentiating \( f \) with respect to \( y \).
The steps for partial differentiation here include:
- Identifying the function and variable of differentiation
- Applying necessary differentiation rules
- Simplifying the result
\[ \frac{-x^2 - xy}{(y^2 - x^2)\text{sqrt}(y^2 - x^2)} \].Understanding these techniques build a foundation for tackling complex multivariable calculus problems effectively.
Other exercises in this chapter
Problem 13
In Exercises 13 through 16, prove that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) exists. \(f(x, y)=\frac{x y}{\sqrt{x^{2}+y^{2}}}\)
View solution Problem 14
In Exercises 11 through 14 , find the total derivative \(d u / d t\) by two methods: (a) Use the chain rule; (b) make the substitutions for \(x\) and \(y\) or f
View solution Problem 14
In Exercises 13 through 16, prove that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) exists. \(f(x, y)=\frac{x^{3}+y^{3}}{x^{2}+y^{2}}\)
View solution Problem 15
In Exercises 15 through 18, show that \(u(x, y)\) satisfies the equation $$ \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 $$ whi
View solution