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Find the differentials of the following functions: a) \(z=\frac{x}{y}\) b) \(z=\log \sqrt{x^{2}+y^{2}}\) c) \(z=\frac{x y}{1-x-y}\) d) \(z=(x-2 y)^{5} e^{x \mathrm{x}}\) e) \(z=\arctan \frac{y}{x}\) f) \(u=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\)

Short Answer

Expert verified
Answer: $$dz = \frac{1}{y} dx - \frac{x}{y^2} dy$$ b) What is the differential of \(z\) with respect to \(x\) and \(y\) for the function \(z=\log \sqrt{x^{2}+y^{2}}\)? Answer: $$dz = \frac{x}{x^2+y^2} dx + \frac{y}{x^2+y^2} dy$$ c) What is the differential of \(z\) with respect to \(x\) and \(y\) for the function \(z=\frac{x y}{1-x-y}\)? Answer: $$dz = \frac{y-y^2}{(1-x-y)^2} dx + \frac{x-x^2}{(1-x-y)^2} dy$$

Step by step solution

01

a) \(z=\frac{x}{y}\)

First, we need to find the partial derivatives of \(z\) with respect to \(x\) and \(y\). For \(\frac{\partial z}{\partial x}\): \(\frac{\partial}{\partial x}\left(\frac{x}{y}\right) = \frac{y}{y^2} = \frac{1}{y}\) For \(\frac{\partial z}{\partial y}\): \(\frac{\partial}{\partial y}\left(\frac{x}{y}\right) = -\frac{x}{y^2}\) The differential of \(z\) with respect to \(x\) and \(y\) is: $$dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy = \frac{1}{y} dx - \frac{x}{y^2} dy$$
02

b) \(z=\log \sqrt{x^{2}+y^{2}}\)

First, we will simplify the function as follows: \(z = \log(x^{2}+y^{2})^{\frac{1}{2}} = \frac{1}{2}\log(x^{2}+y^{2})\) Now, we will find the partial derivatives of \(z\) with respect to \(x\) and \(y\). For \(\frac{\partial z}{\partial x}\): \(\frac{\partial}{\partial x}\left(\frac{1}{2}\log(x^2+y^2)\right) = \frac{1}{2}\cdot\frac{1}{x^2+y^2}\cdot2x = \frac{x}{x^2+y^2}\) For \(\frac{\partial z}{\partial y}\): \(\frac{\partial}{\partial y}\left(\frac{1}{2}\log(x^2+y^2)\right) = \frac{1}{2}\cdot\frac{1}{x^2+y^2}\cdot2y = \frac{y}{x^2+y^2}\) The differential of \(z\) with respect to \(x\) and \(y\) is: $$dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy = \frac{x}{x^2+y^2} dx + \frac{y}{x^2+y^2} dy$$
03

c) \(z=\frac{x y}{1-x-y}\)

We will compute the partial derivatives of \(z\) with respect to \(x\) and \(y\). For \(\frac{\partial z}{\partial x}\): \(\frac{\partial}{\partial x}\left(\frac{xy}{1-x-y}\right) = \frac{y(1-x-y)-xy(-1)}{(1-x-y)^2} = \frac{y-y^2}{(1-x-y)^2}\) For \(\frac{\partial z}{\partial y}\): \(\frac{\partial}{\partial y}\left(\frac{xy}{1-x-y}\right) = \frac{x(1-x-y)-xy(-1)}{(1-x-y)^2} = \frac{x-x^2}{(1-x-y)^2}\) The differential of \(z\) with respect to \(x\) and \(y\) is: $$dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy = \frac{y-y^2}{(1-x-y)^2} dx + \frac{x-x^2}{(1-x-y)^2} dy$$

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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 an extension of the regular derivatives you learn in calculus, tailored to functions with multiple variables. When you have a function like \(z = \frac{x}{y}\), we can think of it as a surface over the \(xy\)-plane. Each point \((x, y)\) on this plane maps to a unique value \(z\). To understand how the function changes as you move along one variable while keeping others constant, you compute the partial derivative.For instance, if you take the partial derivative of \(z = \frac{x}{y}\) with respect to \(x\), you treat \(y\) as a constant. The resulting partial derivative, \(\frac{1}{y}\), tells us the rate at which \(z\) changes with \(x\) alone. Similarly, taking the derivative with respect to \(y\) yields \(-\frac{x}{y^2}\), indicating how \(z\) evolves with changes in \(y\).Partial derivatives are fundamental because they allow us to understand the geometry and behavior of multivariable functions. By combining partial derivatives, we acquire differentials, which express how the entire function shifts due to small changes in all input variables.
Chain Rule
The chain rule is a tool in calculus for calculating the derivative of composite functions. In multivariable calculus, the chain rule extends to handle scenarios where functions are composed with one or more variables themselves dependent on other variables.Suppose you have a composite function like \(z=\log\sqrt{x^2+y^2}\). At first glance, it looks complicated, but we can simplify it to \(z = \frac{1}{2}\log(x^2+y^2)\). When finding partial derivatives, you treat the inner function \((x^2+y^2)\) separately and apply the chain rule sequentially.The partial derivative with respect to \(x\) demands you differentiate \(\frac{1}{2} \log(x^2+y^2)\) concerning \(x\). Using the chain rule, you first differentiate \(\log(x^2+y^2)\) in terms of its argument \((x^2 + y^2)\), then multiply by the derivative of \((x^2 + y^2)\) with respect to \(x\). This layered differentiation allows for precise calculations of the derivative in context.The chain rule is robust, aiding in tackling nested functions arising in real-world and numerous mathematical applications.
Multivariable Calculus
Multivariable calculus builds on concepts from single-variable calculus, extending them to functions with more than one variable. It is vital for solving problems where outcomes depend on multiple factors.Take the function \(z=\frac{xy}{1-x-y}\), defined in two variables. Analyzing its behavior requires partial derivatives, providing insights into changes across two dimensions.The function's geometry becomes critical, especially when examining surfaces represented by \(z=f(x,y)\). Multivariable calculus helps in understanding how slopes and curves develop, and it allows for the examination of rates of change along multiple axes.Furthermore, multivariable calculus includes various calculation techniques, such as finding differentials (like \(dz\)), vectors, and matrices, facilitating the exploration of tangent planes to surfaces. All these tools are essential for delving into more complex fields such as physics, engineering, economics, and beyond, where systems rarely behave in a single dimension.
Logarithmic Differentiation
Logarithmic differentiation is a method used for finding derivatives of functions that are products or quotients of other functions, or where applying other tools like the power rule becomes cumbersome.Take \(z = \log \sqrt{x^2 + y^2}\) as an example. By rewriting this expression, we transform the function into \(z = \frac{1}{2}\log(x^2 + y^2)\). This form is more straightforward to deal with due to the properties of logarithms such as \(\log(ab) = \log(a) + \log(b)\) and \(\log(a^b) = b\log(a)\).In logarithmic differentiation, you differentiate logarithmic expressions after applying these properties, which simplifies finding partial derivatives.This method is particularly beneficial when encountering functions with variables in both the base and the exponent or complex products. It facilitates the calculation by breaking them into manageable pieces and is a powerful tool in both single-variable and multivariable calculus contexts.

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Most popular questions from this chapter

Determine the points \((x, y)\) and directions for which the directional derivative of \(u=\) \(3 x^{2}+y^{2}\) has its largest value, if \((x, y)\) is restricted to lie on the circle \(x^{2}+y^{2}=1\).

A function \(f(x, y)\), defined for all \((x, y)\), satisfies the conditions $$ f(x, 0)=\sin x, \quad \frac{\partial f}{\partial y} \equiv 0 . $$ Evaluate \(f(\pi / 2,2), f(\pi, 3), f(x, 1)\).

For each of the following surfaces, find the tangent plane and normal line at the point indicated. verifying that the point is in the surface: a) \(x^{2}+y^{2}+z^{2}=9\) at \((2,2,1)\) b) \(e^{v^{2}+v^{2}}-z^{2}=0\) at \((0,0,1)\) c) \(x^{3}-x y^{2}+y z^{2}-z^{3}=0\) at \((1,1,1)\) d) \(x^{2}+y^{2}-z^{2}=0\) at \((0,0,0)\) Why does the procedure break down in (d)? Show by graphing that a solution is impossible. e) \(x y-z=0\) at \(\left(x_{1}, y_{1}, z_{1}\right)\), where \(x_{1} y_{1}=z 1\) f) \(x y+y z+x z=1\) at \(\left(x_{1}, y_{1} z_{1}\right)\), where \(x_{1} y_{1}+y_{1} z_{1}+x_{1} z_{1}=1\)

Verify that the mixed derivatives are identical for the following cases: a) \(\frac{\partial^{2} z}{\partial x \partial y}\), and \(\frac{\partial^{2} z}{\partial y \partial x}\) for \(z=\frac{x}{x^{2}+y^{2}}\), b) \(\frac{\partial^{3} w}{\partial x \partial y \partial z}, \frac{\partial^{3} w}{\partial z \partial y \partial x}\), and \(\frac{\partial^{3} w}{\partial y \partial z \partial x}\) for \(w=\sqrt{x^{2}+y^{2}+z^{2}}\).

Let \(z=f(x, y)=x y /\left(x^{2}+y^{2}\right)\) except at \((0,0)\); let \(f(0,0)=0\). Show that \(d z / \partial x\) and \(d=/ d y\) exist for all \((x, y)\) and are continuous except at \((0,0)\). Show by the Fundamental Lemma that z has a differential for \((x, y) \neq(0,0)\) but not at \((0,0)\), since \(f\) is discontinuous at \((0,0)\). IIt is instructive to graph the level curves of \(f\). I

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