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Chapter 3: Problem 24

Use implicit differentiation to find \(d y / d x\) and then \(d^{2} y / d x^{2} .\) Write the solutions in terms of \(x\) and \(y\) only. $$y^{2}-2 x=1-2 y$$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{1}{y+1} \) and \( \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^3} \).

Step by step solution

01

Differentiate both sides with respect to x

First, start by differentiating both sides of the equation \(y^2 - 2x = 1 - 2y\) with respect to \(x\). Note that \(y\) is a function of \(x\), so use the chain rule on \(y^2\) and \(-2y\). The differentiation gives: \[ \frac{d}{dx}(y^2) - \frac{d}{dx}(2x) = \frac{d}{dx}(1) - \frac{d}{dx}(2y) \] Which simplifies to: \[ 2y \frac{dy}{dx} - 2 = -2 \frac{dy}{dx} \]
02

Solve for dy/dx

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \). Add \(2 \frac{dy}{dx}\) to both sides: \[ 2y \frac{dy}{dx} + 2 \frac{dy}{dx} = 2 \] Factor out \( \frac{dy}{dx} \) from the left side: \[ (2y + 2)\frac{dy}{dx} = 2 \] Divide both sides by \(2y + 2\): \[ \frac{dy}{dx} = \frac{2}{2y + 2} = \frac{1}{y + 1} \]
03

Find the second derivative d虏y/dx虏

Differentiate \( \frac{dy}{dx} = \frac{1}{y+1} \) again with respect to \(x\) to find \( \frac{d^2y}{dx^2} \). Using the quotient rule and chain rule: \[ \frac{d}{dx}\left(\frac{1}{y+1}\right) = -\frac{1}{(y+1)^2} \times \frac{d}{dx}(y+1) \] Which simplifies to: \[ \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^2} \times \frac{dy}{dx} \] Substitute \( \frac{dy}{dx} = \frac{1}{y+1} \) from Step 2: \[ \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^2} \times \frac{1}{y+1} = -\frac{1}{(y+1)^3} \]
04

Summary of the results

The first derivative \( \frac{dy}{dx} \) in terms of \(x\) and \(y\) is \( \frac{dy}{dx} = \frac{1}{y+1} \). The second derivative \( \frac{d^2y}{dx^2} \) is \( \frac{d^2y}{dx^2} = -\frac{1}{(y+1)^3} \). These results satisfy the conditions given in the exercise and are correct in terms of \(x\) and \(y\).

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Key Concepts

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

Derivatives
In calculus, a derivative represents how a function changes as its input changes. It is essentially the slope of the function at any given point. When dealing with a problem that requires implicit differentiation, like the one we discussed, you're often trying to find the derivative of an equation where both the dependent and independent variables are intertwined.
  • Derivative notation: We often use \( \frac{dy}{dx}\) to show how \( y \) changes with respect to \( x \). Here, \( y\) is seen as a function of \( x \), even if not explicitly.
  • Process: You differentiate both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \), hence the need for using implicit differentiation.
  • Solution obtained: Through the problem-solving steps provided, the goal was to find \(\frac{dy}{dx}\) in terms of \( y\), leading us to an expression that is independent of \( x \) but involves both \( y \) and constants.
Understanding derivatives in this context shows their importance in describing how changes in one variable affect another through a linked relationship.
Chain Rule
The chain rule is an essential tool in calculus for differentiating composite functions. When you have a composite function, such as \( y(x) \) where \( y \) is a function dependent on another function \( x \), you apply the chain rule to appropriately differentiate.
  • How it works: If you have a function \( z = f(g(x)) \), the chain rule states that the derivative of \( z \) with respect to \( x \) is \( \frac{dz}{dx} = f'(g(x)) \cdot g'(x) \).
  • Application in implicit differentiation: In our problem, terms such as \( y^2 \) require using the chain rule. By considering \( y \) as a function of \( x \), we differentiate \( y^2 \) as \( 2y \cdot \frac{dy}{dx} \).
By recognizing and employing the chain rule, you can effectively handle changes that involve multiple related functions within equations.
Quotient Rule
The quotient rule comes into play when differentiating a function that is the quotient of two other functions. It is especially handy for tasks similar to our initial derivative-differentiating step.
  • Rule definition: If you have a function \( h(x) = \frac{f(x)}{g(x)} \), its derivative is given by the formula: \( \frac{dh}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).
  • Usage: For more complex fractions, the quotient rule simplifies finding the dy/dx that includes the ratio of derivatives.
  • Finding the second derivative: In our problem, while differentiating \(\frac{1}{y+1}\) to find the second derivative \(\frac{d^2y}{dx^2}\), the quotient rule was crucial. Using the rule ensures that the sequential derivatives respect their functional relationships.
Utilizing the quotient rule allows you to untangle and compute derivatives of functions interrelated through division, critical for problems involving multiple derivatives.

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Most popular questions from this chapter

Find the values of a. \(\sec ^{-1} 1.5\) b. \(\csc ^{-1}(-1.5)\) c. \(\cot ^{-1} 2\)

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and the length \(L\) of a simple pendulum with the equation $$T=2 \pi \sqrt{\frac{L}{g}}$$ where \(g\) is the constant acceleration of gravity at the pendulum's location. If we measure \(g\) in centimeters per second squared, we measure \(L\) in centimeters and \(T\) in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to \(L\). In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$\frac{d L}{d u}=k L$$ Assuming this to be the case, show that the rate at which the period changes with respect to temperature is \(k T / 2\)

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on Earth's surface, depending on the change in \(g\). By keeping track of \(\Delta T\), we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\) a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001\) sec. Find \(d g\) and estimate the value of g at the new location.

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=\frac{3 x+2}{2 x-11}, \quad-2 \leq x \leq 2, \quad x_{0}=1 / 2$$

Use logarithmic differentiation or the method in Example 7 to find the derivative of \(y\) with respect to the given independent variable. $$y=(x+1)^{x}$$
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