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$$ \begin{array}{l}{15-18(\text { a) Find the differential dy and (b) evaluate dy for the }} \\ {\text { given values of } x \text { and } d x \text { . }}\end{array} $$ $$ y=1 /(x+1), \quad x=1, \quad d x=-0.01 $$

Short Answer

Expert verified
For \(x = 1\) and \(dx = -0.01\), \(dy = 0.0025\).

Step by step solution

01

Find the Derivative

The first step in finding the differential \(dy\) is to find the derivative of the function \(y = \frac{1}{x+1}\). Using the quotient rule is unnecessary here, so instead, we rewrite the function as \(y = (x+1)^{-1}\) and use the power rule for differentiation. The derivative \( \frac{dy}{dx} \) is given by \( -1(x+1)^{-2} = \frac{-1}{(x+1)^2} \).
02

Calculate the Differential dy

Now that we have the derivative, we can find the differential \(dy\) using the formula \( dy = \frac{dy}{dx} \cdot dx \). Substituting the derivative we found in Step 1, we get \( dy = \frac{-1}{(x+1)^2} \cdot dx \).
03

Evaluate dy at Given x and dx

To evaluate \(dy\), substitute \(x = 1\) and \(dx = -0.01\) into the expression for \(dy\). This leads to \(dy = \frac{-1}{(1+1)^2} \cdot (-0.01) = \frac{-1}{4} \cdot (-0.01) = 0.0025\).

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Key Concepts

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

Derivative Calculation
In differential calculus, calculating a derivative is like finding how a function changes as its input changes slightly. For the exercise provided, we want to calculate the derivative of the function \( y = \frac{1}{x+1} \). To simplify the process, we can rewrite this as \( y = (x+1)^{-1} \). Calculating the derivative involves using techniques that determine the rate of change. By rewriting the function, we make it easier to apply the power rule (which we'll explain later) instead of dealing with quotients. After applying the method of differentiation, we find that the derivative \( \frac{dy}{dx} \) is \(-\frac{1}{(x+1)^2}\). This means for every tiny change in \(x\), we can calculate how much \(y\) is expected to change, which is exactly what a derivative tells us.
Differential Equations
Differential equations involve equations that relate a function with its derivatives. The exercise here only involves basic differential methods, calculating what's called a simple differential \( dy \). A differential like \( dy \) helps us approximate how much \( y \) changes when \( x \) is changed by a small amount, \( dx \). To compute \( dy \):
  • We multiply the derivative \( \frac{dy}{dx} \) by the small change in \(x\), \( dx \).
  • This yields \( dy = \frac{-1}{(x+1)^2} \cdot dx \).
Thus, given an initial value for \( x \) and a tiny adjustment \( dx \), we can estimate the resultant change in \( y \). Such approximation techniques are useful across many scientific fields when solving problems relating to how variables interact.
Power Rule for Differentiation
One of the key skills in differential calculus is knowledge of differentiation rules, including the power rule. It’s a straightforward method to find the derivative of polynomial-like terms. The power rule states: if \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \). In our exercise, the original function \( y = (x+1)^{-1} \) is in a form that allows application of the power rule. By setting the exponent to \(-1\), we can directly find its derivative. Applying the rule, the exponent \(-1\) is multiplied by the original function, reducing the power by one and giving us \( -1(x+1)^{-2} \). Knowing this rule allows us to compute derivatives swiftly without cumbersome algebra like dealing with fractions unless necessary.

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Most popular questions from this chapter

At noon, ship \(A\) is 100 \(\mathrm{km}\) west of ship \(\mathrm{B} .\) Ship \(\mathrm{A}\) is sailing south at 35 \(\mathrm{km} / \mathrm{h}\) and ship \(\mathrm{B}\) is sailing north at 25 \(\mathrm{km} / \mathrm{h}\) . How fast is the distance between the ships changing at \(4 : 00 \mathrm{PM}\) ?

Use the Chain Rule to show that if \(\theta\) is measured in degrees, then $$\frac { d } { d \theta } ( \sin \theta ) = \frac { \pi } { 180 } \cos \theta$$ This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: The differentiation formulas would not be as simple if we used degree measure.)

$$ \begin{array}{l}{\text { The radius of a circular disk is given as } 24 \mathrm{cm} \text { with a maxi- }} \\ {\text { mum error in measurement of } 0.2 \mathrm{cm} .} \\ {\text { (a) Use differentials to estimate the maximum error in the }} \\ {\text { calculated area of the disk. }} \\ {\text { (b) What is the relative error? What is the percentage error? }}\end{array} $$

Use the definition of derivative to prove that \(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1\)

$$ \begin{array}{c}{\text { When blood flows along a blood vessel, the flux F (the }} \\ {\text { volume of blood per unit time that flows past a given point) }} \\ {\text { is proportional to the fourth power of the radius R of the }} \\ {\text { blood vessel: }} \\ {\quad F=k R^{4}}\end{array} $$ This is known as Poiseuille's Law; we will show why it is true in Section 8.4 . ) A partially clogged artery can be expanded by an operation called angioplasty, in which a balloon-tipped catheter is inflated inside the artery in order to widen it and restore the normal blood flow. Show that the relative change in \(F\) is about four times the relative change in R. How will a 5\(\%\) increase in the radius affect the flow of blood?

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