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

Correct Answer is : − 4x(1+x2)2-\;4x/(1+x²)²−(1+x2)24x

Correct Answer is : − 4x(1+x2)2-\;4x/(1+x²)²−(1+x2)24x

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TG EAMCET 4 May 2025 Shift 1 Paper

Question : 65 of 160

Marks: +1, -0

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

\;\frac{dy}{dx}=

Solution:

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

\;\; \text{ Put } \;\; \frac{1+x^2}{2}=\cos^2(2\theta)

\;\;\;\;\; x^2=2\cos^2(2\theta)-1=\cos4

\;\; \text{ Also, } \; 0\le\frac{1+x^2}{2}\le1

\;\;\; 0\le1+x^2\le2

\;\Rightarrow\;\; x^2\le1\Rightarrow-1\le x\le1

\;\Rightarrow\;\; -1\le\cos4\theta\le1

\;\Rightarrow\;\; 0\le4\theta\le\pi\Rightarrow0\le\theta\le\frac{\pi}{4}

Differentiating with resped to

x

\;\Rightarrow\;\; 2x\;\frac{dx}{d\theta}=-4\sin4\theta

\;\Rightarrow\;\; \frac{dx}{d\theta}=\frac{-2\sin4\theta}{\sqrt{\cos4\theta}}\;\;\;\cdots(ii)

and

y=\tan^2\cos^{-1}(\cos2\theta)=\tan^2(2\theta)

\Rightarrow\;\; \frac{dy}{d\theta}=2\tan(2\theta)\times\sec^2(2\theta)\times2\ldots\;\text{(i)}\;

\Rightarrow\;\; \frac{dy}{d\theta}=\frac{4\sin(2\theta)}{\cos^3(2\theta)}

\Rightarrow\;\; \frac{dy}{dx}=\frac{4\sin(2\theta)\times\sqrt{\cos4\theta}}{\cos^3(2\theta)(-4)\sin2\theta\cdot\cos2\theta}

\;=-\;\frac{\sqrt{\cos4\theta}}{\cos^4(2\theta)}

\;=\;\frac{-\sqrt{x^2}}{\left(\frac{1+\cos4\theta}{2}\right)^2}=\frac{-4x}{(1+x^2)^2}

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