Skip to main content
Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.8.14b
Chapter 3, Problem 3.8.14b

13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
x = e^y; (2, ln 2)

Verified step by step guidance
1
Start by recognizing that the equation given is in implicit form: \( x = e^y \). This means that \( y \) is not isolated on one side of the equation.
To find the derivative \( \frac{dy}{dx} \), apply implicit differentiation to both sides of the equation with respect to \( x \). Differentiate \( x \) to get 1, and differentiate \( e^y \) using the chain rule to get \( e^y \cdot \frac{dy}{dx} \).
Set up the equation from the differentiation: \( 1 = e^y \cdot \frac{dy}{dx} \).
Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation: \( \frac{dy}{dx} = \frac{1}{e^y} \).
Substitute the given point \((2, \ln 2)\) into the expression for \( \frac{dy}{dx} \). Since \( y = \ln 2 \), calculate \( e^{\ln 2} \) which simplifies to 2, and find the slope at this point.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m

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 curves defined by equations that cannot be easily rearranged.
Recommended video:
Guided course
05:14
Finding The Implicit Derivative

Slope of a Curve

The slope of a curve at a given point represents the rate of change of the function at that point, which is mathematically defined as the derivative of the function. For a curve defined implicitly, the slope can be found by evaluating the derivative obtained through implicit differentiation. The slope is often denoted as 'dy/dx' and indicates how steep the curve is at the specified coordinates.
Recommended video:
11:41
Summary of Curve Sketching

Exponential and Natural Logarithm Functions

Exponential functions, such as e^y, and natural logarithm functions, like ln(x), are fundamental in calculus. The function e^y is the inverse of ln(y), and they exhibit unique properties, such as the derivative of e^y being e^y dy/dx. Understanding these functions is crucial for evaluating expressions and derivatives involving exponential growth or decay, especially when working with implicit relationships.
Recommended video:
05:18
Derivative of the Natural Logarithmic Function
Related Practice
Textbook Question

A race Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions θ(t) and ϕ(t), respectively, where 0≤t≤4 and t is measured in minutes (see figure). These angles are measured in radians, where θ=ϕ=0 represent the starting position and θ=ϕ=2π represent the finish position. The angular velocities of the runners are θ′(t) and ϕ′(t). <IMAGE>

b. Which runner has the greater average angular velocity?

1
views
Textbook Question

Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

<IMAGE>

b. h′(2)h^{\(\prime\)}\(\left\)(2\(\right\))

Textbook Question

Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = √3x; a= 12

1
views
Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

cos y = x; (0, π/2)

Textbook Question

{Use of Tech} Spring oscillations A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance x (in inches) of the mass from its equilibrium position after t seconds is given by the function x(t) = 10sin t - 10cos t, where x is positive when the mass is above the equilibrium position. <IMAGE>

b. Find dx/dt and interpret the meaning of this derivative.  

1
views
Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(x) = 1/x+1; a = -1/2;5