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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 3.10.13
Chapter 3, Problem 3.10.13

Evaluate the derivative of the following functions.
f(x) = sin-1 2x

Verified step by step guidance
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Step 1: Recognize that \( f(x) = \sin^{-1}(2x) \) is the inverse sine function, also known as arcsine, applied to \( 2x \).
Step 2: Recall the derivative formula for the inverse sine function: \( \frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \), where \( u \) is a function of \( x \).
Step 3: Identify \( u = 2x \) in this problem, so you need to find \( \frac{du}{dx} \).
Step 4: Differentiate \( u = 2x \) with respect to \( x \), which gives \( \frac{du}{dx} = 2 \).
Step 5: Substitute \( u = 2x \) and \( \frac{du}{dx} = 2 \) into the derivative formula: \( \frac{d}{dx}[\sin^{-1}(2x)] = \frac{1}{\sqrt{1-(2x)^2}} \cdot 2 \).

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Key Concepts

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

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In calculus, the derivative is often denoted as f'(x) or df/dx, and it provides critical information about the function's behavior, such as its slope and points of tangency.
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Inverse Trigonometric Functions

Inverse trigonometric functions, such as sin<sup>-1</sup>(x), are the functions that reverse the action of the standard trigonometric functions. For example, sin<sup>-1</sup>(x) gives the angle whose sine is x. These functions have specific domains and ranges, and their derivatives can be derived using implicit differentiation or known derivative formulas, which are essential for evaluating derivatives involving these functions.
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Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It states that if a function y = f(g(x)) is composed of two functions, the derivative can be found by multiplying the derivative of the outer function f with the derivative of the inner function g. This rule is particularly useful when dealing with functions like sin<sup>-1</sup>(2x), where the inner function (2x) must be differentiated alongside the outer function (sin<sup>-1</sup>), requiring careful application of the chain rule.
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Related Practice
Textbook Question

Derivatives Find and simplify the derivative of the following functions.

h(x) = (x−1)(2x²-1) / (x³-1)

Textbook Question

75–86. Logarithmic differentiation Use logarithmic differentiation to evaluate f'(x).

f(x) = x^In x

Textbook Question

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

x2+93\(\sqrt\)[3]{x^2+9}

Textbook Question

At all times, the length of the long leg of a right triangle is 3 times the length x of the short leg of the triangle. If the area of the triangle changes with respect to time t, find equations relating the area A to x and dA/dt to dx/dt.

Textbook Question

Using identities Use the identity sin 2x=2 sin x cos x sin 2 to find d/dx (sin 2x). Then use the identity cos 2x = cos² x−sin² x to express the derivative of sin 2x in terms of cos 2x.

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Textbook Question

Calculate the derivative of the following functions. In some cases, it is useful to use the properties of logarithms to simplify the functions before computing f'(x).

f(x) = In(3x + 1)⁴