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Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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use another way to prove the function.

diferen Differentiate with respect to x. derivative of y = sinx follows by differentiating both sides of x = sin y sin x is the value of y such that x=, cOs Inverse Sine and Its Derivative differen (Exerci Recall from Section 1.4 that y sin-1 x is the value of y such that -T/2<y < /2. The domain of sinx is {x:-1 <x derivative of y = sin- x follows by differentiating both sides of ure to x, simplifying, and solving for dy/dx: -1 %3D x = sin y y = sinx x = sin y x X = sin y d (x) = (sin y) %3D dx dx 1 = (cos y) dy Chain Rule on the right side dx dy 1 dy Solve for dx %3D dx cos y The identity sin? y + cos? y = 1 is used to express this derivative in te Solving for cos y yields cos y = ±V1 – sin y x = sin y x = sin' y x sin y x² = sin y %3D .2 = ±V1 - x?. %3D | Because y is restricted to the interval -T/2 < y < /2, we have cos y 2 0. The we choose the positive branch of the square root, and it follows that dy d. 1 (sinx) dx VI - 1 x2 dx
Transcribed Image Text:diferen Differentiate with respect to x. derivative of y = sinx follows by differentiating both sides of x = sin y sin x is the value of y such that x=, cOs Inverse Sine and Its Derivative differen (Exerci Recall from Section 1.4 that y sin-1 x is the value of y such that -T/2<y < /2. The domain of sinx is {x:-1 <x derivative of y = sin- x follows by differentiating both sides of ure to x, simplifying, and solving for dy/dx: -1 %3D x = sin y y = sinx x = sin y x X = sin y d (x) = (sin y) %3D dx dx 1 = (cos y) dy Chain Rule on the right side dx dy 1 dy Solve for dx %3D dx cos y The identity sin? y + cos? y = 1 is used to express this derivative in te Solving for cos y yields cos y = ±V1 – sin y x = sin y x = sin' y x sin y x² = sin y %3D .2 = ±V1 - x?. %3D | Because y is restricted to the interval -T/2 < y < /2, we have cos y 2 0. The we choose the positive branch of the square root, and it follows that dy d. 1 (sinx) dx VI - 1 x2 dx
d - 1 Prove that for the inverse cosine function, (cos-1 x) = for - 1<x< 1. dx 1-x2 Jse a proof analogous to the proof of Theorem 3.21 (P. 234).
Transcribed Image Text:d - 1 Prove that for the inverse cosine function, (cos-1 x) = for - 1<x< 1. dx 1-x2 Jse a proof analogous to the proof of Theorem 3.21 (P. 234).
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Step 1

Here, the objective is to prove the following:

 ddxcos-1x=-11-x2   -1<x<1

 

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