If y=Tan-1sinx+Cosec-1e2x+1 then dydx=

y=tan-1sinx+cosec-1e2x+1 =11+sin2x cosx·12x-2e4x+2-1 =cosx2x1+sin2x -2e4x+2-1

If y=Tan-1sinx+Cosec-1e2x+1 then dydx=

y=tan-1sinx+cosec-1e2x+1 =11+sin2x cosx·12x-2e4x+2-1 =cosx2x1+sin2x -2e4x+2-1

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If $y = {Tan}^{- 1} (\sin \sqrt{x}) + {Cosec}^{- 1} (e^{2 x + 1})$ then $\frac{dy}{dx} =$

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$\frac{1}{\sqrt{x} (1 + \sin^{2} \sqrt{x})} + \frac{1}{\sqrt{e^{4 x + 2} + 1}}$

$\frac{\cos \sqrt{x}}{2 \sqrt{x} (1 + \sin^{2} \sqrt{x})} + \frac{2}{\sqrt{e^{4 x + 2} - 1}}$

$\frac{\cos \sqrt{x}}{(1 + \sin^{2} \sqrt{x})} + \frac{2}{\sqrt{e^{4 x + 2} + 1}}$

$\frac{1}{2 \sqrt{x}} \frac{\cos \sqrt{x}}{(1 + \sin^{2} \sqrt{x})} - \frac{2}{\sqrt{e^{2 x + 1} - 1}}$

answer is D.

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Detailed Solution

$y = \tan^{- 1} (\sin \sqrt{x}) + {cosec}^{- 1} (e^{2 x + 1}) = \frac{1}{(1 + \sin^{2} \sqrt{x})} \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} - \frac{2}{\sqrt{e^{4 x + 2} - 1}} = \frac{\cos \sqrt{x}}{2 \sqrt{x} (1 + \sin^{2} \sqrt{x})} - \frac{2}{\sqrt{e^{4 x + 2} - 1}}$