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Chapter 3: Problem 16

Differentiate the functions and find the slope of the tangent line at the given value of the independent variable. $$y=\frac{x+3}{1-x}, \quad x=-2$$

Short Answer

Expert verified
The slope of the tangent line at \( x = -2 \) is \( \frac{4}{9} \).

Step by step solution

01

Rewrite the Function

First, write the given function in a form that's suitable for using the quotient rule for differentiation. The function given is \( y = \frac{x+3}{1-x} \). This is already in a simple fraction form suitable for the quotient rule.
02

Apply the Quotient Rule

To differentiate \( y = \frac{x+3}{1-x} \), use the quotient rule: if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). - Here, \( u = x+3 \) and \( v = 1-x \).- \( \frac{du}{dx} = 1 \).- \( \frac{dv}{dx} = -1 \).Applying the rule:\[ \frac{dy}{dx} = \frac{(1-x) \cdot 1 - (x+3) \cdot (-1)}{(1-x)^2} \] which simplifies to:\[ \frac{dy}{dx} = \frac{1 - x + x + 3}{(1-x)^2} = \frac{4}{(1-x)^2} \].
03

Plug in the Given Value

Substitute \( x = -2 \) into the derivative \( \frac{dy}{dx} = \frac{4}{(1-x)^2} \):\[ \frac{dy}{dx} \bigg|_{x = -2} = \frac{4}{(1 - (-2))^2} = \frac{4}{3^2} = \frac{4}{9} \].
04

Interpret the Result

The slope of the tangent line to the curve at \( x = -2 \) is \( \frac{4}{9} \). This means that for a very small change in \( x \) around \( x = -2 \), the change in \( y \) is approximately \( \frac{4}{9} \) times the change in \( x \).

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Key Concepts

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

Quotient Rule
The quotient rule is an essential tool in calculus when dealing with the differentiation of functions that are expressed as a ratio of two functions. For any function written as \( y = \frac{u}{v} \), the derivative can be found using the rule:
  • \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)
This formula allows us to find the rate at which \( y \) changes with respect to \( x \), even when \( y \) is a complex fraction.
Consider the function \( y = \frac{x+3}{1-x} \). Here, \( u = x+3 \) and \( v = 1-x \). The derivatives of \( u \) and \( v \) are \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -1 \) respectively.
By plugging these into the formula, one can find the derivative of the function. This allows us to understand how steep or gentle the curve is at any particular point.
Tangent Line
A tangent line to a curve at a given point is a straight line that just "grazes" the curve at that point. In calculus, the concept of a tangent line plays a crucial role as it reflects the behavior of the curve around that specific location.
For a function, the equation of a tangent line at a point \( x = a \) can be found using the derivative:
  • The slope of the tangent line is \( m = \frac{dy}{dx} \bigg|_{x = a} \).
For the given problem, evaluating the derivative \( \frac{dy}{dx} = \frac{4}{(1-x)^2} \) at \( x = -2 \) tells us that the slope of the tangent is \( \frac{4}{9} \).
This means that at the point where \( x = -2 \), the tangent line has a consistent slope, providing instant insight into how the function changes at that point.
Slope Calculation
Calculating the slope of a function at a specific point is fundamental for understanding the direction and steepness of the curve at that spot. This slope matches the concept of the derivative at that point.
In our example, the slope of the tangent line at \( x = -2 \) is calculated as \( \frac{4}{9} \).
  • This calculation comes from substituting the specific \( x \) value into the derived expression for \( \frac{dy}{dx} \).
By using the formula \( \frac{4}{(1-x)^2} \), and substituting \( x = -2 \), we calculate:
  • \( \frac{dy}{dx} \bigg|_{x = -2} = \frac{4}{3^2} = \frac{4}{9} \)
Thus, this slope tells us that for very tiny shifts in \( x \) around \( -2 \), the function \( y \) changes at a rate proportional to \( \frac{4}{9} \), which can be a critical factor in analyzing functionality and behavior.

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Most popular questions from this chapter

What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$f(h)=\frac{\sin h}{h}$$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$\lim _{h \rightarrow 0} \frac{\cos h-1}{h}$$ c. Now go back to the derivation of the formula for the derivative of \(\sin x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of \(\cos x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of \(\sin x\) and \(\cos x ?\)

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

On a morning of a day when the sun will pass directly overhead, the shadow of an \(24 \mathrm{m}\) building on level ground is \(18 \mathrm{m}\) long. At the moment in question, the angle \(\theta\) the sun makes with the ground is increasing at the rate of \(0.27^{\circ} / \mathrm{min} . \mathrm{At}\) what rate is the shadow decreasing? (Remember to use radians. Express your answer in centimeters per minute, to the nearest tenth.)

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x \sqrt{1+2 y}+y=x^{2}, \quad P(1,0)$$

A light shines from the top of a pole \(15 \mathrm{m}\) high. A ball is dropped from the same height from a point \(9 \mathrm{m}\) away from the light. (See accompanying figure.) How fast is the shadow of the ball moving along the ground \(1 / 2\) s later? (Assume the ball falls a distance \(s=4.9 t^{2} \mathrm{m}\) in \(t\) seconds.)
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