Textbook Question
Evaluate the integrals in Exercises 33–54.
∫(from ln3 to ln2) (e^x) dx
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.6.128
128. Derive the formula dy/dx = 1/(1+x²) for the derivative of y = arctan(x) by differentiating both sides of the equivalent equation tan(y)=x.
Verified step by step guidance1
Start with the given equivalent equation: \(\tan(y) = x\). This relates \(y\) and \(x\) implicitly.
Differentiate both sides of the equation with respect to \(x\). Use implicit differentiation on the left side: \(\frac{d}{dx}[\tan(y)] = \frac{d}{dx}[x]\).
Apply the chain rule to the left side: \(\sec^2(y) \cdot \frac{dy}{dx} = 1\), since the derivative of \(\tan(y)\) with respect to \(y\) is \(\sec^2(y)\) and then multiply by \(\frac{dy}{dx}\).
Solve for \(\frac{dy}{dx}\) by dividing both sides by \(\sec^2(y)\): \(\frac{dy}{dx} = \frac{1}{\sec^2(y)}\).
Use the trigonometric identity \(\sec^2(y) = 1 + \tan^2(y)\) and substitute \(\tan(y) = x\) to rewrite the derivative as \(\frac{dy}{dx} = \frac{1}{1 + x^2}\).
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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 find the derivative of a function defined implicitly rather than explicitly. When y is given in terms of x through an equation like tan(y) = x, we differentiate both sides with respect to x, treating y as a function of x and applying the chain rule.
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Finding The Implicit Derivative
Derivative of the Tangent Function
The derivative of tan(y) with respect to y is sec²(y). This fact is essential when differentiating tan(y) implicitly, as it allows us to relate dy/dx to the derivative of tan(y) with respect to x by using the chain rule.
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05:13Slopes of Tangent Lines
Trigonometric Identity for sec²(y)
The identity sec²(y) = 1 + tan²(y) helps simplify the derivative expression. Since tan(y) = x, substituting this into the identity allows us to express sec²(y) in terms of x, which is crucial for isolating dy/dx and deriving the formula dy/dx = 1/(1 + x²).
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7:17Verifying Trig Equations as Identities
Related Practice
Textbook Question
In Exercises 59–86, find the derivative of y with respect to the given independent variable.
77. y = log₃(((x + 1)/(x − 1))^(ln 3))
Textbook Question
73. Find the area between the curves y=ln(x) and y=ln(2x) from x=1 to x=5.
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Textbook Question
Evaluate the integrals in Exercises 39–56.
49. ∫3sec²t/(6 + 3tan(t)) dt
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Textbook Question
In Exercises 73 and 74, repeat the steps above to solve for the functions y=f(x) and x=f^(-1)(y) defined implicitly by the given equations over the interval.
73. y^(1/3) - 1 = (x+2)³, -5 ≤ x ≤ 5, x_0 = -3/2
Textbook Question
Evaluate the integrals in Exercises 53–76.
73. ∫(from 0 to ln√3) e^x dx/(1+e^(2x))