/*! 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 98 If \(x=\sin ^{-1}\left(\frac{3 \... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 98

If \(x=\sin ^{-1}\left(\frac{3 \sin t+4 \cos t}{5}\right)\) and \(y=\sin ^{-1}\left(\frac{6 \cos t+8 \sin t}{10}\right)\), find \(\frac{d y}{d x}\).

Short Answer

Expert verified
The required derivative \( \frac{dy}{dx} \) is \( \frac{-6\sin t+8\cos t}{-3\cos t+4\sin t} \) once simplified.

Step by step solution

01

Find the Derivative of x with Respect to t

Differentiate \(x=\sin ^{-1}\left(\frac{3 \sin t+4 \cos t}{5}\right) \) with respect to t using the chain rule. This yields: \( \frac{dx}{dt} = \frac{1}{\sqrt{1 - \left( \frac{3 \sin t + 4 \cos t}{5} \right)^2}} \cdot \frac{d}{dt} \left( \frac{3 \sin t + 4 \cos t}{5} \right) \). By simplifying, we get: \( \frac{dx}{dt}=\frac{-3\cos t+4\sin t}{5\sqrt{1 - \left( \frac{3 \sin t + 4 \cos t}{5} \right)^2}} \).
02

Find the Derivative of y with Respect to t

Similarly, differentiate \(y=\sin ^{-1}\left(\frac{6 \cos t+8 \sin t}{10}\right)\) with respect to t. This gives us: \( \frac{dy}{dt} = \frac{1}{\sqrt{1 - \left( \frac{6 \cos t + 8 \sin t}{10} \right)^2}} \cdot \frac{d}{dt} \left( \frac{3 \cos t - 4 \sin t}{10} \right) \). After simplifying, we obtain: \( \frac{dy}{dt}=\frac{-6\sin t+8\cos t}{10\sqrt{1 - \left( \frac{6 \cos t + 8 \sin t}{10} \right)^2}} \).
03

Compute the Derivative of y with Respect to x

Now that we have both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), we can use the relation \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) to find the required derivative. Plugging in the earlier obtained values, we get \( \frac{dy}{dx}=\frac{\frac{-6\sin t+8\cos t}{10\sqrt{1 - \left( \frac{6 \cos t + 8 \sin t}{10} \right)^2}}}{\frac{-3\cos t+4\sin t}{5\sqrt{1 - \left( \frac{3 \sin t + 4 \cos t}{5} \right)^2}}} \) then simplifying this yields the final answer.

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Most popular questions from this chapter

If \(y=\left(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\right)\), prove that \(2 x y \frac{d y}{d x}=\frac{x}{a}-\frac{a}{x}\)

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

If \(x^{2}-y^{2}=t-\frac{1}{t}\) and \(x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}\) then prove that \(x^{3} y \frac{d y}{d x}+1=0\)

Differentiate \(\tan ^{-1}\left\\{\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right\\}\) w.r.t. \(\cos ^{-1} x^{2}\).

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