Problem 5
Question
Find the first partial derivatives of the function. $$ g(u, v)=\frac{u^{3}+v^{3}}{u^{2}+v^{2}} $$
Step-by-Step Solution
Verified Answer
The partial derivatives are \( \frac{\partial g}{\partial u} = \frac{u^4 + 3u^2v^2 - 2uv^3}{(u^2 + v^2)^2} \) and \( \frac{\partial g}{\partial v} = \frac{3v^2u^2 + v^4 - 2u^3v}{(u^2 + v^2)^2} \).
1Step 1: Understand the Function
We have the function \( g(u, v) = \frac{u^3 + v^3}{u^2 + v^2} \). The task is to find the partial derivatives \( \frac{\partial g}{\partial u} \) and \( \frac{\partial g}{\partial v} \).
2Step 2: Identify the Quotient Rule
Since the function is a quotient, we need to apply the quotient rule to find the partial derivatives. Recall the quotient rule: for \( f(x) = \frac{N(x)}{D(x)} \), its derivative is \( f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{D(x)^2} \).
3Step 3: Compute \( \frac{\partial g}{\partial u} \)
- Set \( N(u, v) = u^3 + v^3 \), a function of \( u \) only.- Set \( D(u, v) = u^2 + v^2 \), a function of \( u \) only.- Derivative of \( N \) with respect to \( u \) is \( \frac{\partial}{\partial u} (u^3 + v^3) = 3u^2 \).- Derivative of \( D \) with respect to \( u \) is \( \frac{\partial}{\partial u} (u^2 + v^2) = 2u \).Apply quotient rule:\[ \frac{\partial g}{\partial u} = \frac{(3u^2)(u^2 + v^2) - (u^3 + v^3)(2u)}{(u^2 + v^2)^2} \]Simplify:\[ = \frac{3u^2u^2 + 3u^2v^2 - 2u^4 - 2uv^3}{(u^2 + v^2)^2} \]\[ = \frac{u^4 + 3u^2v^2 - 2uv^3}{(u^2 + v^2)^2} \].
4Step 4: Compute \( \frac{\partial g}{\partial v} \)
- Set \( N(v, u) = u^3 + v^3 \), considering \( v \) only.- Set \( D(v, u) = u^2 + v^2 \), considering \( v \) only.- Derivative of \( N \) with respect to \( v \) is \( \frac{\partial}{\partial v} (u^3 + v^3) = 3v^2 \).- Derivative of \( D \) with respect to \( v \) is \( \frac{\partial}{\partial v} (u^2 + v^2) = 2v \).Apply quotient rule:\[ \frac{\partial g}{\partial v} = \frac{(3v^2)(u^2 + v^2) - (u^3 + v^3)(2v)}{(u^2 + v^2)^2} \]Simplify:\[ = \frac{3v^2u^2 + 3v^4 - 2u^3v - 2v^4}{(u^2 + v^2)^2} \]\[ = \frac{3v^2u^2 + v^4 - 2u^3v}{(u^2 + v^2)^2} \].
Key Concepts
quotient rulefunctions of two variablesderivative simplification
quotient rule
The quotient rule is an essential tool when dealing with derivatives of division forms involving variable terms. Consider the function given in the exercise, \( g(u, v) = \frac{u^3 + v^3}{u^2 + v^2} \). Since it is in the form of a fraction with a numerator and a denominator, the quotient rule is perfect for finding the derivative.
The quotient rule states that if you have a function \( f(x) = \frac{N(x)}{D(x)} \), the derivative is:
In this problem, it helps to treat \( u \) and \( v \) independently as you look for their respective partial derivatives. The applications of the quotient rule result in complex fractions that often require further simplification to get to the final form.
The quotient rule states that if you have a function \( f(x) = \frac{N(x)}{D(x)} \), the derivative is:
- \( f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{D(x)^2} \)
In this problem, it helps to treat \( u \) and \( v \) independently as you look for their respective partial derivatives. The applications of the quotient rule result in complex fractions that often require further simplification to get to the final form.
functions of two variables
Functions of two variables involve expressions like \( z = f(x, y) \) or \( g(u, v) \), where the output depends on two different inputs. In our exercise, the function \( g(u, v) = \frac{u^3 + v^3}{u^2 + v^2} \) illustrates this concept perfectly.
With functions of two variables, you can consider each variable separately when finding partial derivatives. This means you treat one variable as a constant while differentiating with respect to the other. For instance, when finding \( \frac{\partial g}{\partial u} \), you view \( v \) as a constant and vice versa.
Partial derivatives tell us how a function changes as one of the input variables is varied, while the others are held constant, a concept critical in multivariable calculus. Understanding these changes helps in many fields, including physics and economics, where variables influence results differently.
With functions of two variables, you can consider each variable separately when finding partial derivatives. This means you treat one variable as a constant while differentiating with respect to the other. For instance, when finding \( \frac{\partial g}{\partial u} \), you view \( v \) as a constant and vice versa.
Partial derivatives tell us how a function changes as one of the input variables is varied, while the others are held constant, a concept critical in multivariable calculus. Understanding these changes helps in many fields, including physics and economics, where variables influence results differently.
derivative simplification
Derivative simplification is the process of reducing complex derivative expressions to their simplest form. In this exercise, after applying the quotient rule to find the partial derivatives \( \frac{\partial g}{\partial u} \) and \( \frac{\partial g}{\partial v} \), you will obtain equations that involve several terms. Simplification involves:
In our problem, for \( \frac{\partial g}{\partial u} \), you simplify:
- Combining like terms
- Reducing fractions
In our problem, for \( \frac{\partial g}{\partial u} \), you simplify:
- \( \frac{3u^2 (u^2 + v^2) - (u^3 + v^3) (2u)}{(u^2 + v^2)^2} \)
- \( \frac{u^4 + 3u^2v^2 - 2uv^3}{(u^2 + v^2)^2} \)
Other exercises in this chapter
Problem 5
Find the gradient of the function. $$ f(x, y, z)=2 x^{2}-y^{2}-4 z^{2} $$
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Compute \(d z / d t\). $$ z=\sqrt{2 x-4 y} ; x=\ln t, y=1-3 t^{3} $$
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Evaluate the limit. $$ \lim _{(x, y, z) \rightarrow(2,1,-1)} \frac{2 x^{2} y-x z^{2}}{y^{2}-x z} $$
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Find the domain of the function. \(g(x, y)=\sqrt{x^{2}+y^{2}-25}\)
View solution