/*! 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 129 If \(y=\sin ^{-1} x\), prove tha... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 129

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

Short Answer

Expert verified
After step-by-step verification, this exercise proves that for \(y=\sin ^{-1} x\), \(\left(1-x^{2}\right) y_{2}-x y_{1}=0\) is indeed valid.

Step by step solution

01

Find the first and second derivative of inverse sine function

By rule of differentiation, the first derivative of \(y = \sin ^{-1} x\) is given by, \(y_{1} = dy/dx = 1 / \sqrt{1 - x^2}\). Now for second derivative, first be recognized that the derivative of \(\sqrt{1 - x^2}\) is \(-x / \sqrt{1 - x^2}\). Then, by the quotient rule of differentiation, we find that \(y_{2} = d^2y/dx^2 = -x/ (1 - x^2 )^{3/2}\).
02

Substitution into the given equation

Now, substituting our found values of \(y_{1}\) and \(y_{2}\) back to the given equation \(\left(1-x^{2}\right) y_{2}-x y_{1}\), we have: \(\left(1-x^{2}\right) (-x/ (1 - x^2 )^{3/2}) - x (1 / \sqrt{1 - x^2})\)
03

Simplify the equation

Simplify above step and it gives \(-x + x =0\)
04

Proving equation validity

As our calculation shows, \(\left(1-x^{2}\right) y_{2}-x y_{1}=0\) holds true, we therefore have proved that this equation is correct based on the given condition, \(y=\sin ^{-1} x\)

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Most popular questions from this chapter

For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)

If \(y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \text { to } \infty}}}\) then find \(\frac{d y}{d x}\).

If \(a, b, c\) be non-zero real numbers such that $$ \begin{aligned} & \int_{0}^{1}\left(1+\cos ^{8} x\right)\left(a x^{2}+b x+c\right) d x \\ =& \int_{0}^{2}\left(1+\cos ^{8} x\right)\left(a x^{2}+b x+c\right) d x \end{aligned} $$ then the equation \(a x^{2}+b x+c=0\) will have a root between (a) \((1,3)\) (b) \((1,2)\) (c) \((2,3)\) (d) \((3,1)\)

If \(x \sqrt{1+y}+y \sqrt{1+x}=0\), prove that \(\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}\)

Find \(\frac{d y}{d x}\), if \(2 x^{2}+3 x y+3 y^{2}=1\)
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