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Chapter 7: Problem 50

Find \(d x / d y\). $$ y=x-2 / x \text { for } x<0 $$

Short Answer

Expert verified
The derivative \( \frac{dx}{dy} = \frac{1}{1 + \frac{2}{x^2}} \).

Step by step solution

01

Express y in terms of x

Given the equation, rearrange it to explicitly write the expression for \(y\) in terms of \(x\):\[ y = x - \frac{2}{x} \]
02

Differentiate y with respect to x

To find \( \frac{dy}{dx} \), differentiate the expression obtained in step 1 with respect to \(x\): \[ \frac{dy}{dx} = \frac{d}{dx} \left(x - \frac{2}{x}\right). \] Use the derivative rules: the derivative of \(x\) is 1, and the derivative of \(-\frac{2}{x}\) using the power rule is \(\frac{2}{x^2}\). Thus, \[ \frac{dy}{dx} = 1 + \frac{2}{x^2}. \]
03

Find dx/dy

To find \( \frac{dx}{dy} \), we need to find the reciprocal of \( \frac{dy}{dx} \). Hence,\[ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{1 + \frac{2}{x^2}}. \]

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Key Concepts

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

Understanding Derivatives
Derivatives are a core concept in calculus, representing the rate at which a function changes. In more intuitive terms, if you imagine a curve on a graph, the derivative at a specific point tells you how steep the curve is at that point. This steepness is essentially the slope of the tangent line to the curve.
  • The derivative of a function is commonly denoted by \( \frac{dy}{dx} \), representing the change in \( y \) with respect to the change in \( x \).
  • To compute derivatives, we use various rules such as the power rule, product rule, and chain rule, depending on the form of the function.
  • The result of a derivative function provides insights into increasing or decreasing behavior of the original function, concavity, and inflection points.
Differentiating requires acute attention to detail, ensuring each component of the function is handled per its rules.
Exploring Reciprocal Functions
A reciprocal function is a specific type of function where each output value is the reciprocal (or multiplicative inverse) of the input. Mathematically, if you have a function like \( f(x) = \frac{1}{x} \), it is called a reciprocal function.
  • Reciprocal functions lead to interesting behaviors, especially when it comes to differentiation, as it often involves transforming the exponent in the denominator.
  • In the context of our exercise, after differentiating \( y \) with respect to \( x \) and obtaining \( \frac{dy}{dx} \), finding the reciprocal gives us \( \frac{dx}{dy} \), which is essentially flipping the derivative to yield a new perspective on the rate of change.
  • Understanding reciprocal relationships helps unravel complex expressions in calculus and solve differential equations.
This flipping or inverting process highlights the symmetrical nature of rates of change between the two variables.
Applying the Power Rule
The power rule is a straightforward yet powerful tool in calculus for differentiating functions of the form \( x^n \). According to the power rule, if \( f(x) = x^n \), then the derivative \( \frac{df}{dx} \) is \( nx^{n-1} \).
  • This rule dramatically simplifies the differentiation process, especially when dealing with polynomial functions.
  • When applying the power rule to our exercise's function \( y = x - \frac{2}{x} \), remember that \( \frac{2}{x} \) can be rewritten as \( 2x^{-1} \), making differentiation straightforward using the power rule, resulting in \( \frac{2}{x^2} \) for that term.
  • The power rule is not just limited to whole numbers but also applies to negative and fractional exponents, broadening its applications across various calculus problems.
Its simplicity and robustness make the power rule indispensable for anyone learning calculus.

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Most popular questions from this chapter

Prove that $$ \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \quad \text { for } \quad x y \neq 1 $$ provided that the value of the expression on the left-hand side lies in \((-\pi / 2, \pi / 2)\).

Find $$ \lim _{x \rightarrow 0} \frac{\int_{0}^{x}(x-t) \sin \left(t^{2}\right) d t}{\ln \left(1+x^{4}\right)} $$

Find the general solution of the linear differential equation. $$ \frac{d y}{d x}-y=\frac{1}{1-e^{-x}} $$

Find the area \(A\) of the region bounded by the graphs of \(y=e^{2 x}\) and \(y=e^{-2 x}\) and the line \(x=1 / 2\).

A right triangle \(T\) in the first quadrant has legs on the axes and hypotenuse tangent to the curve \(y=e^{-x}\). a. Show that for any \(\varepsilon>0\) the area of \(T\) is less than \(\varepsilon\) if the base is sufficiently large. b. Show that the area of \(T\) is maximal if the base of \(T\) has length \(2 .\)
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