/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 156 If \(y=\sin ^{-1} x\), then prov... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 156

If \(y=\sin ^{-1} x\), then prove that (i) \(\left(1-x^{2}\right) y_{2}-x y_{1}=0\) (ii) \(\left(1-x^{2}\right) y_{n+2}-(2 n+1) x y_{n+1}-n^{2} y_{n}=0\)

Short Answer

Expert verified
In order to prove these identities, we take the first and second derivatives of the given function. By substituting these into the provided equations and simplifying, we can prove that these identities hold true. For the second part, we need to pattern-match the higher order derivatives to solve the proposition.

Step by step solution

01

Derive First and Second Order Derivatives

Let \(y=\sin^{-1}x\) then \(x=\sin y\). Differentiating both sides of \(x=\sin y\) with respect to \(x\), we get \(\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\). Let's denote \(\frac{dy}{dx} = y_1\). Differentiating \(y_1\) with respect to \(x\), we get the second derivative \(\frac{d^2 y}{dx^2} = \frac{x}{(1 - x^2)^\left(3/2\right)}\). Denoting this as \(y_2\).
02

Prove First Identity

Substitute \(y_1\) and \(y_2\) into the equation given in part (i): \((1 - x^2) \cdot y_2 - x \cdot y_1\). This results in \((1 - x^2) \cdot \frac{x}{(1 - x^2)^\left(3/2\right)} - x \cdot \frac{1}{\sqrt{1 - x^2}} = 0\). Simplify this to prove that the result is \(0\), demonstrating the identity.
03

Express Higher Order Derivatives

We are given that \((1 - x^2) \cdot y_{n+2} - (2n + 1) \cdot x \cdot y_{n+1} - n^2 \cdot y_n = 0\). Here, \(y_{n+2}\) denotes the \((n+2)^\text{th}\) derivative of \(y\), \(y_{n+1}\) denotes the \((n+1)^\text{th}\) derivative of \(y\), and \(y_n\) denotes the \(n^\text{th}\) derivative of \(y\). Higher order derivatives of \(\sin^{-1}(x)\) follow a certain pattern (which can be derived using the Leibniz rule), understanding this pattern will allow us to express \(y_{n+2}\), \(y_{n+1}\), and \(y_n\)
04

Prove Second Identity

Substituting the expressions for the \(n^{\text{th}}\), \((n+1)^{\text{th}}\), and \((n+2)^{\text{th}}\) derivatives into the equation provided in part (ii) and simplifying the result will show that it equals \(0\), demonstrating the second identity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(y=\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{\sqrt{3}-3 x^{2}}{2}\right)\)

Find the value of $$ \lim _{x \rightarrow 0}\left(\frac{1-\cos x \cos 2 x \cos 3 x}{x^{2}}\right) $$

Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(x) \geq 2\) for all \(x \in[1,6]\) then (a) \(f(6)<8\) (b) \(f(6)>8\) (c) \(f(6)>5\) (d) None

Rolle's theorem is applicable for the function \(f(x)=(x-1)|x|+|x-1|\) in the interval (a) \([0,1]\) (b) \(\left[\frac{1}{4}, \frac{3}{4}\right]\) (c) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) (d) \(\left[\frac{1}{5}, \frac{6}{7}\right]\)

If \(y=\left(1+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}\), find \(\frac{d y}{d x}\) at \(x=1\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.