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Partial derivatives Find the first partial derivatives of the following functions. $$f(w, z)=\frac{w}{w^{2}+z^{2}}$$

Short Answer

Expert verified
Answer: The first partial derivatives of the given function are: $$\frac{\partial}{\partial w} f(w, z) = \frac{-w^2+z^2}{(w^2+z^2)^2}$$ $$\frac{\partial}{\partial z} f(w, z) = \frac{-2wz}{(w^2+z^2)^2}$$

Step by step solution

01

Finding the partial derivative with respect to \(w\)

To find the partial derivative of \(f(w, z)\) with respect to \(w\), we differentiate the function with respect to \(w\) while keeping \(z\) constant. We will use the quotient rule to differentiate the function: $$\frac{\partial}{\partial w} f(w, z) = \frac{\mathrm{d}(\frac{w}{w^2 + z^2})}{\mathrm{d} w} = \frac{(w^2+z^2)\times\frac{d w}{d w} - w\times\frac{d(w^2+z^2)}{d w}}{(w^2+z^2)^2}$$ Next, we differentiate the terms in the numerator with respect to \(w\).
02

Differentiating the terms in the numerator

Now, we differentiate the terms in the numerator with respect to \(w\): $$\frac{d w}{d w} = 1$$ $$\frac{d(w^2+z^2)}{d w} = 2w$$ Substitute these derivatives back into the expression for the partial derivative: $$\frac{\partial}{\partial w} f(w, z) = \frac{(w^2+z^2)\times 1 - w\times 2w}{(w^2+z^2)^2}$$ Simplify the expression: $$\frac{\partial}{\partial w} f(w, z) = \frac{w^2+z^2 - 2w^2}{(w^2+z^2)^2} = \frac{-w^2+z^2}{(w^2+z^2)^2}$$
03

Finding the partial derivative with respect to \(z\)

To find the partial derivative of \(f(w, z)\) with respect to \(z\), we differentiate the function with respect to \(z\) while keeping \(w\) constant. We will use the same quotient rule as before: $$\frac{\partial}{\partial z} f(w, z) = \frac{\mathrm{d}(\frac{w}{w^2 + z^2})}{\mathrm{d} z} = \frac{(w^2+z^2)\times\frac{d w}{d z} - w\times\frac{d(w^2+z^2)}{d z}}{(w^2+z^2)^2}$$ Next, we differentiate the terms in the numerator with respect to \(z\).
04

Differentiating the terms in the numerator

Now, we differentiate the terms in the numerator with respect to \(z\): $$\frac{d w}{d z} = 0$$ $$\frac{d(w^2+z^2)}{d z} = 2z$$ Substitute these derivatives back into the expression for the partial derivative: $$\frac{\partial}{\partial z} f(w, z) = \frac{(w^2+z^2)\times 0 - w\times 2z}{(w^2+z^2)^2}$$ Simplify the expression: $$\frac{\partial}{\partial z} f(w, z) = \frac{-2wz}{(w^2+z^2)^2}$$ In conclusion, we have found the first partial derivatives of the given function: $$\frac{\partial}{\partial w} f(w, z) = \frac{-w^2+z^2}{(w^2+z^2)^2}$$ $$\frac{\partial}{\partial z} f(w, z) = \frac{-2wz}{(w^2+z^2)^2}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
When faced with a function in the form of a quotient—a ratio of two functions—one useful tool for differentiation is the quotient rule. It is essential in single-variable calculus, and just as much in multivariable calculus. The quotient rule states that the derivative of \( \frac{u(v)}{v(w, z)} \) with respect to a variable, say \( w \), is given by \( \frac{v(w, z)\frac{d u}{d w} - u(v)\frac{d v}{d w}}{[v(w, z)]^2} \).
Understanding the quotient rule allows us to tackle the differentiation of rational functions, where the numerator and the denominator are both functions of the variable with respect to which we're differentiating. It's important to apply the rule correctly by differentiating the numerator and the denominator separately and then combining the results as per the formula.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. In fields such as physics, engineering, and economics, variables often depend on multiple factors, which is where multivariable calculus becomes essential. Instead of dealing with a single input and output, we consider functions with inputs like \( w \) and \( z \), and how they change with respect to each variable. Concepts such as partial derivatives, gradient vectors, and multiple integrals are fundamental in studying the surfaces, solids, and change rates in multivariable systems.
Differentiation
Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value is changing at a particular point. In the context of multivariable calculus, differentiation is performed with respect to each variable independently while holding the other variables constant. This gives rise to partial derivatives, which represent the rate of change of a multivariable function with respect to one variable at a time.
Partial Derivative with Respect to w
A partial derivative with respect to a variable, such as \( w \), is denoted as \( \frac{\partial}{\partial w} \) and is calculated by differentiating the function concerning \( w \) while treating all other variables as constants. In our exercise, the differentiation of \( \frac{w}{w^2 + z^2} \) with respect to \( w \) involves using the quotient rule, treating \( z \) as a constant. This results in \( \frac{\partial}{\partial w} f(w, z) = \frac{-w^2+z^2}{(w^2+z^2)^2} \) after substituting the differentiated terms and simplifying. It tells us how the function \( f \) changes as \( w \) varies, with \( z \) fixed.
Partial Derivative with Respect to z
Similarly, the partial derivative with respect to \( z \) is represented as \( \frac{\partial}{\partial z} \) and finds the rate of change of the function when \( z \) varies and \( w \) is held constant. Applying the quotient rule in our example gives \( \frac{\partial}{\partial z} f(w, z) = \frac{-2wz}{(w^2+z^2)^2} \) after differentiating the numerator where \( w \) is considered constant and \( z \) is the variable of interest. This partial derivative highlights how changes in \( z \) affect the value of our function \( f \) while \( w \) remains unchanged.

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Most popular questions from this chapter

Travel cost The cost of a trip that is \(L\) miles long, driving a car that gets \(m\) miles per gallon, with gas costs of \(\$ p /\) gal is \(C=L p / m\) dollars. Suppose you plan a trip of \(L=1500 \mathrm{mi}\) in a car that gets \(m=32 \mathrm{mi} / \mathrm{gal},\) with gas costs of \(p=\$ 3.80 / \mathrm{gal}\) a. Explain how the cost function is derived. b. Compute the partial derivatives \(C_{L}, C_{m^{\prime}}\) and \(C_{p^{\prime}}\). Explain the meaning of the signs of the derivatives in the context of this problem. c. Estimate the change in the total cost of the trip if \(L\) changes from \(L=1500\) to \(L=1520, m\) changes from \(m=32\) to \(m=31,\) and \(p\) changes from \(p=\$ 3.80\) to \(p=\$ 3.85\) d. Is the total cost of the trip (with \(L=1500 \mathrm{mi}, m=32 \mathrm{mi} / \mathrm{gal}\). and \(p=\$ 3.80\) ) more sensitive to a \(1 \%\) change in \(L,\) in \(m,\) or in \(p\) (assuming the other two variables are fixed)? Explain.

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The planes tangent to the cylinder \(x^{2}+y^{2}=1\) in \(R^{3}\) all have the form \(a x+b z+c=0\) b. Suppose \(w=x y / z,\) for \(x>0, y>0,\) and \(z>0 .\) A decrease in \(z\) with \(x\) and \(y\) fixed results in an increase in \(w\) c. The gradient \(\nabla F(a, b, c)\) lies in the plane tangent to the surface \(F(x, y, z)=0\) at \((a, b, c)\)

Find the absolute maximum and minimum values of the following functions over the given regions \(R .\) \(f(x, y)=2 x^{2}-4 x+3 y^{2}+2\) \(R=\left\\{(x, y):(x-1)^{2}+y^{2} \leq 1\right\\}\) (This is Exercise 51 Section \(15.7 .)\)

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