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Chapter 2: Problem 29

Finding the Slope of a Graph In Exercises \(25-32,\) find \(d y / d x\) by implicit differentiation. Then find the slope of the graph at the given point. \((x+y)^{3}=x^{3}+y^{3}, \quad(-1,1)\)

Short Answer

Expert verified
The slope of the graph at point \((-1, 1)\) is \(1\).

Step by step solution

01

Differentiate the equation implicitly

Differentiate both sides of the equation \((x+y)^{3}=x^{3}+y^{3}\) with respect to \(x\). Use the chain rule to differentiate \( (x + y)^3 \) which gives \(3(x+y)^2\left(1 + \frac{dy}{dx}\right)\) and the right side differentiates to \(3x^2 + 3y^2 \frac{dy}{dx}\).
02

Rearrange for dy/dx

Rearrange the equation to solve for \(\frac{dy}{dx}\). You end up with \(\frac{dy}{dx} = \frac{3x^2 - 3(x+y)^2}{3(y+y^2) - 3(x+y)^2}\).
03

Substituting given point

Substitute \((-1,1)\) for \((x, y)\) in the expression of \(\frac{dy}{dx}\) we got from Step 2, which leads to \(\frac{dy}{dx} = \frac{3(-1)^2 - 3((-1)+1)^2}{3(1+1^2) - 3((-1)+1)^2} = 1 \).

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Key Concepts

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

Chain Rule in Implicit Differentiation
When we want to find derivatives involving both \(x\) and \(y\), especially when \(y\) isn't isolated, we use implicit differentiation. Here, the chain rule is crucial. The chain rule is a way of differentiating composite functions, and it basically tells us how to find the derivative of a "function within a function."

In our exercise, we start with \((x + y)^3\). Using the chain rule:
  • The outer function is \(t^3\), and its derivative is \(3t^2\).
  • The inner function \(t = x + y\) needs to differentiate with respect to \(x\) to get \(1 + \frac{dy}{dx})\).
This results in \(3(x+y)^2(1 + \frac{dy}{dx})\).
The whole process helps us handle equations where one variable is inside another, linking the changes of \(x\) and \(y\).
Understanding the Slope of a Graph
The slope of a graph at a specific point is a measure of how steep the graph is at that location. In mathematics, the slope is usually represented by \(\frac{dy}{dx}\), which tells us how much \(y\) changes for a small change in \(x\).

Finding the slope at a point involves determining the rate of change around that point. If the slope is positive, the graph rises as \(x\) increases, and if it's negative, the graph falls.

In practical terms, finding this slope through implicit differentiation helps us understand curves that cannot be expressed as \(y = f(x)\). This exercise showed how the slope is calculated using \(x\) and \(y\) derivatives simultaneously.
The Process of Solving for dy/dx
After differentiating the equation implicitly, the next task is to solve for \(\frac{dy}{dx}\). This involves rearranging the differentiated equation to isolate \(\frac{dy}{dx}\) on one side of the equation.

In our solution, we are rearranging the terms from the differentiated part which involved getting all \(\frac{dy}{dx}\) terms on one side:
  • Set up: Combine terms to segregate \(\frac{dy}{dx}\).
  • Simplify: Group like terms and factor out \(\frac{dy}{dx}\) when needed.
  • Result: You鈥檒l get the equation in the form of \(\frac{dy}{dx} = ...\).
This process is essential because it transforms the implicit equation into an explicit expression for the derivative.
Substitute Given Points
Once we have \(\frac{dy}{dx}\), substituting specific \((x, y)\) values helps us find the slope at that particular point on the graph.

With the exercise鈥檚 given point \((-1,1)\), substituting these values involves inserting into the expression for \(\frac{dy}{dx}\) we derived:
  • Replace \(x = -1\) and \(y = 1\).
  • Calculate to find \(\frac{dy}{dx}\) at these coordinates.
  • Here, the substitution gave us \(\frac{dy}{dx} = 1\), indicating the slope at point \((-1, 1)\).
This substitution helps pinpoint differences at specific locations on the graph, providing insights into the graph's behavior at precise spots.

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Most popular questions from this chapter

pumped into the trough. Moving Ladder A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

Writing Use a graphing utility to graph the two functions \(f(x)=x^{2}+1\) and \(g(x)=|x|+1\) in the same viewing window. Use the zoom and trace features to analyze the graphs near the point \((0,1)\) . What do you observe? Which function is differentiable at this point? Write a short paragraph describing the geometric significance of differentiability at a point.

Finding a Pattern Consider the function \(f(x)=\sin \beta x\) where \(\beta\) is a constant. (a) Find the first-, second-, third-, and fourth-order derivatives of the function. (b) Verify that the function and its second derivative satisfy the equation \(f^{\prime \prime}(x)+\beta^{2} f(x)=0\) (c) Use the results of part (a) to write general rules for the even- and odd- order derivatives \(f^{(2 k)}(x)\) and \(f^{(2 k-1)}(x)\) [Hint: \((-1)^{k}\) is positive if \(k\) is even and negative if \(k\) is odd. \(]\)

\(\begin{array}{l}{\text { Approximating a Derivative In Exercises } 67 \text { and } 68 \text { , }} \\ {\text { evaluate } f(2) \text { and } f(2.1) \text { and use the results to approximate } f^{\prime}(2) \text { . }}\end{array}\) $$f(x)=\frac{1}{4} x^{3}$$

Tangent Line Find the equation(s) of the tangent line(s) to the graph of the curve \(y=x^{3}-9 x\) through the point \((1,-9)\) not on the graph.
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