If y=^{-1}≤ft\{{x}{1+{1-x²}\} +≤ft\{2^{-1}{1-x/1+x}\}, then dy/dx =

Correct Answer is : 1−2x21−x2{1-2x}{2{1-x²}}21−x21−2x

Correct Answer is : 1−2x21−x2{1-2x}{2{1-x²}}21−x21−2x

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AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Section: Mathematics

Question : 65 of 160

Marks: +1, -0

y=\tan^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\}

+\sin\left\{2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right\}

\frac{dy}{dx}

Solution:

Consider the equation.

y=\tan^{-1} x \left\{\frac{x}{1+\sqrt{1-x^{2}}}\right\}

+\sin \left\{2 \tan^{-1}\sqrt{\frac{1-x}{1+x}}\right\}

Substitute

\cos 2\theta

for

x

in the above equation.

y=\tan^{-1} \left\{ \frac{\cos 2\theta}{1+\sqrt{\sin^{2} 2\theta}} \right\}

+\sin \left\{2 \tan^{-1}\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}\right\}

=\tan^{-1} \left\{\frac{\cos 2\theta}{1+\sin 2\theta}\right\}+\sin \left\{2 \tan^{-1} \sqrt{\frac{2\sin^{2}\theta}{2\cos^{2}\theta}}\right\}

=\tan^{-1} \left\{\frac{\cos^{2} \theta-\sin^{2} \theta}{(\cos \theta+\sin \theta)^{2}}\right\}

+\sin \left\{2 \tan^{-1}(\tan \theta)\right\}

=\tan^{-1} \left\{\frac{\cos \theta-\sin \theta}{(\cos \theta+\sin \theta)}\right\}+\sin \{2 \theta\}

Solve further,

y=\tan^{-1} \left\{\frac{1-\tan \theta}{1+\tan \theta}\right\}+\sin 2 \theta

y=\tan^{-1} \left(\tan \left(\frac{\pi}{4}-\theta\right) \right)+\sin 2 \theta

= \frac{\pi}{4}-\theta+\sin 2 \theta

= \frac{\pi}{4}- \frac{1}{2} \cos^{-1} x+ \sqrt {1-x^{2}}

Differentiate both side with respect to

x

\frac{d y}{d x}= \frac{1}{2} \cdot \frac{1}{\sqrt{1-x^{2}}}

+ \frac{1}{2 \sqrt{1-x^{2}}}(-2 x)

=\frac{1-2x}{2\sqrt{1-x^{2}}}

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