/*! 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 61 If \(\sin y=x \sin (a+y)\), prov... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 61

If \(\sin y=x \sin (a+y)\), prove that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)

Short Answer

Expert verified
\(\frac{d y}{d x}=\frac{\sin^{2}(a+y)}{\sin a}\)

Step by step solution

01

Differentiate both sides with respect to \(x\).

Since the equation \(\sin y=x \sin (a+y)\) is given, differentiating both sides with respect to \(x\), gives us: \(\cos y \frac{dy}{dx} = \sin(a+y) + x \cos(a+y) \frac{dy}{dx}\).
02

Rearrange the terms to solve for \(\frac{dy}{dx}\).

Rearranging the equation to solve for \(\frac{dy}{dx}\), you have: \(\frac{dy}{dx} = \frac{\sin(a+y)}{\cos y - x \cos(a+y)}\).
03

Apply trigonometric identities.

For the denominator, use the Cosine of Difference Identity, \(\cos(A-B)=\cos A \cos B+\sin A \sin B\), where \(A=y\) and \(B=a+y\). Hence, \(\cos y - x \cos(a+y) = \cos y - x(\cos y \cos a + \sin y \sin a) = (1 - x \cos a) \cos y - x \sin a \sin y\). Then use the original equation \(\sin y=x \sin (a+y)\) to substitute \(\sin y\) in terms of \(x\). Finally, simplify to have the denominator as \((1 - x \cos a) x \sin (a+y)\). Hence, \(\frac{dy}{dx} = \frac{\sin (a+y)}{(1 - x \cos a) x \sin (a+y)} = \frac{1}{(1 - x \cos a) x}\).
04

Apply additional trigonometric identities.

Using the sine version of the Pythagorean identity, \(\sin^2a + \cos^2a = 1\), we can express \(\cos a\) in terms of \(\sin a\), which is \(\cos a = \sqrt{1 - \sin^2a}\). Replace \(\cos a\) with this expression in the denominator, we obtain: \(\frac{dy}{dx} = \frac{1}{(1 - x \sqrt{1 - \sin^2a}) x} = \frac{1}{x - x^2 \sqrt{1 - \sin^2a}}\). Now, multiply numerator and denominator by \(\frac{1}{\sqrt{1 - \sin^2a}}\) we get: \(\frac{dy}{dx} = \frac{\frac{1}{\sqrt{1 - \sin^2a}}}{x \frac{1}{\sqrt{1 - \sin^2a}} - x^2} = \frac{\sin a}{x \sin a - x^2}\). This further simplifies to \(\frac{dy}{dx} = \frac{\sin a}{\sin a(x - x^2)} = \frac{\sin a}{x \sin a(1 - x)} = \frac{1}{x(1 - x)}\), given that \(\sin a \neq 0\).
05

Re-arrange terms.

Rearranging the equation, our solution becomes: \(\frac{dy}{dx} = \frac{1}{x - x^2} = \frac{1}{x(1 - x)} = \frac{\sin^2(a+y)}{\sin a}\), which completes the proof.

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Most popular questions from this chapter

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 \(x=\tan ^{-1}\left(\frac{\sqrt{1+\sin t}+\sqrt{1-\sin t}}{\sqrt{1+\sin t}-\sqrt{1-\sin t}}\right)\) and \(y=\tan ^{-1}\left(\frac{\sqrt{1+t^{2}}-1}{t}\right)\), find \(\frac{d y}{d x}\)

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

If \(x^{y}=e^{x-y}\), prove that, \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^{2}} .\)

There is a polynomial \(P(x)=a x^{3}+b x^{2}+c x+d\) such that \(P(0)=P(1)=-2, P^{\prime}(0)=-1\), then find the value of \(a+b+c+d+10\)
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