Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.8.39

Chapter 3, Problem 3.8.39

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.
√x⁴+y² = 5x+2y³

Verified step by step guidance

Start by differentiating both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, use the chain rule.

Differentiate the left side: For √(x⁴ + y²), use the chain rule. The derivative of √u with respect to u is 1/(2√u), and then multiply by the derivative of the inside function (x⁴ + y²) with respect to x.

Differentiate the right side: For 5x + 2y³, differentiate each term separately. The derivative of 5x with respect to x is 5, and for 2y³, use the chain rule: differentiate y³ with respect to y, which is 3y², and then multiply by dy/dx.

Set the derivatives from both sides equal to each other. This will give you an equation involving dy/dx.

Solve the resulting equation for dy/dx. This will involve isolating dy/dx on one side of the equation, which may require algebraic manipulation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. Instead of solving for one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, applying the chain rule as necessary. This method is particularly useful for equations that are difficult or impossible to rearrange.

Chain Rule

The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. It states that if a function y is defined as a function of u, which in turn is a function of x, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This is essential in implicit differentiation when dealing with terms involving y.

Derivative Notation (dy/dx)

The notation dy/dx represents the derivative of y with respect to x, indicating the rate of change of y as x changes. In the context of implicit differentiation, dy/dx is treated as a variable that can be solved for, even when y is not explicitly defined in terms of x. Understanding this notation is crucial for interpreting the results of differentiation and for solving for the slope of the tangent line at a given point.

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Textbook Question

15–48. Derivatives Find the derivative of the following functions.

P = 40/1+2^-t

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Textbook Question

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas y = cx² form orthogonal trajectories with the family of ellipses x²+2y² = k, where c and k are constants (see figure).

Find dy/dx for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. <IMAGE>

y = cx²; x²+2y² = k, where c and k are constants

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Textbook Question

Vertical tangent lines If a function f is continuous at a and lim x→a| f′(x)|=∞, then the curve y=f(x) has a vertical tangent line at a, and the equation of the tangent line is x=a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 71–72) is used. Use this information to answer the following questions.

Graph the following curves and determine the location of any vertical tangent lines.

a. x²+y² = 9

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Textbook Question

If f is a one-to-one function with f(3)=8 and f′(3)=7, find the equation of the line tangent to y=f^−1(x) at x=8.

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Textbook Question

27–76. Calculate the derivative of the following functions.

$y = {(x^{2} + 2 x + 7)}^{8}$