If y = (sin^{-1} 2x)^2 + (cos^{-1} 2x)^2, the | vedclass

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(C) Given $y = (\sin^{-1} 2x)^2 + (\cos^{-1} 2x)^2$. Using the identity $\sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2}$, we can write $\cos^{-1}

Vedclass · Accepted Answer

$16$

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(C) Given $y = (\sin^{-1} 2x)^2 + (\cos^{-1} 2x)^2$. Using the identity $\sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2}$, we can write $\cos^{-1} 2x = \frac{\pi}{2} - \sin^{-1} 2x$. Substituting this into $y$: $y = (\sin^{-1} 2x)^2 + (\frac{\pi}{2} - \sin^{-1} 2x)^2$ $y = (\sin^{-1} 2x)^2 + \frac{\pi^2}{4} - \pi \sin^{-1} 2x + (\sin^{-1} 2x)^2$ $y = 2(\sin^{-1} 2x)^2 - \pi \sin^{-1} 2x + \frac{\pi^2}{4}$. Differentiating with respect to $x$: $y_1 = 4(\sin^{-1} 2x) \cdot \frac{2}{\sqrt{1-4x^2}} - \pi \cdot \frac{2}{\sqrt{1-4x^2}}$ $y_1 = \frac{8 \sin^{-1} 2x - 2\pi}{\sqrt{1-4x^2}}$. $\sqrt{1-4x^2} y_1 = 8 \sin^{-1} 2x - 2\pi$. Differentiating again with respect to $x$: $\sqrt{1-4x^2} y_2 + y_1 \cdot \frac{-8x}{2\sqrt{1-4x^2}} = 8 \cdot \frac{2}{\sqrt{1-4x^2}}$. Multiply throughout by $\sqrt{1-4x^2}$: $(1-4x^2) y_2 - 4x y_1 = 16$.

If $y = (\sin^{-1} 2x)^2 + (\cos^{-1} 2x)^2$, then $(1 - 4x^2) y_2 - 4x y_1 = $

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