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Chapter 6: Problem 104

If \(y=\cos ^{-1}\left(\frac{5 t+12 \sqrt{1-t^{2}}}{13}\right)\) and \(x=\cos ^{-1}\left(\frac{1-t^{2}}{1+t^{2}}\right)\), find \(\frac{d y}{d x}\)

Short Answer

Expert verified
\(\frac{d y}{d x}\) will be found after substituting the derived \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) values into \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) equation.

Step by step solution

01

Differentiate y with respect to t

Firstly, differentiate y with respect to t: \( \frac{dy}{dt}=-\frac{1}{\sqrt{1-\left(\frac{5t+12\sqrt{1-t^2}}{13}\right)^2}} \cdot \frac{d}{dt}\left(\frac{5t+12\sqrt{1-t^2}}{13}\right) \).
02

Find \(\frac{d}{dt}(\frac{5t+12\sqrt{1-t^2}}{13})\)

Find the derivative of the function in the brackets which is given as \( \frac{5t+12\sqrt{1-t^2}}{13} \). Use the chain rule again to differentiate \( \sqrt{1-t^2} \).
03

Differentiate x with respect to t

Now, differentiate x with respect to t: \( \frac{dx}{dt}=-\frac{2t}{\sqrt{1-\left(\frac{1-t^2}{1+t^2}\right)^2}} \cdot \frac{d}{dt}\left(\frac{1-t^2}{1+t^2}\right) \).
04

Find \(\frac{d}{dt}(\frac{1-t^2}{1+t^2})\)

Differentiate the fraction which is given as \( \frac{1-t^2}{1+t^2} \). Remember to use the quotient rule for differentiation.
05

Find \(\frac{dy}{dx}\)

Finally we have the necessary pieces to find \( \frac{dy}{dx} \). By the chain rule, \( \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \). Substitute your solutions from the previous steps into this formula to find \( \frac{dy}{dx} \).

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Most popular questions from this chapter

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

Find \(\frac{d^{2} y}{d x^{2}}\), if (i) \(x=a t^{2}, y=2 a t\) (ii) \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\) (iii) \(x=a \cos \theta, y=b \sin \theta\)

If \(\tan ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=a\) prove that \(\frac{d y}{d x}=\frac{x(1-\tan a)}{y(1+\tan a)}\).

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If \(y=\cos ^{-1}\left(\frac{5 t+12 \sqrt{1-t^{2}}}{13}\right)\) and \(x=\cos ^{-1}\left(\frac{1-t^{2}}{1+t^{2}}\right)\), find \(\frac{d y}{d x}\)
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