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If y=sin⁡−1xx+1+sec⁡−1x+1xy = \sin^{-1} \sqrt\frac{x}{x+1} + \sec^{-1}\sqrt{\frac{x+1}{x}}, then dydx\frac{dy}{dx} is

Solution

✅ Correct Option: 1

y=sin⁡−1xx+1+sec⁡−1x+1xy = \sin^{-1} \sqrt{\frac{x}{x+1}} + \sec^{-1} \sqrt{\frac{x+1}{x}}


Using the identity sec⁡−1(u)=cos⁡−1(1u)\sec^{-1}(u) = \cos^{-1}\left(\frac{1}{u}\right) with u=x+1xu = \sqrt{\dfrac{x+1}{x}}:

sec⁡−1x+1x=cos⁡−1(1x+1x)\sec^{-1}\sqrt{\frac{x+1}{x}} = \cos^{-1}\left(\frac{1}{\sqrt{\frac{x+1}{x}}}\right)

=cos⁡−1xx+1= \cos^{-1}\sqrt{\frac{x}{x+1}}


y=sin⁡−1xx+1+cos⁡−1xx+1y = \sin^{-1}\sqrt{\frac{x}{x+1}} + \cos^{-1}\sqrt{\frac{x}{x+1}}

Both terms have the same argument t=xx+1t = \sqrt{\dfrac{x}{x+1}}.

Using the identity sin⁡−1(t)+cos⁡−1(t)=π2\sin^{-1}(t) + \cos^{-1}(t) = \dfrac{\pi}{2}, for t∈[−1,1]t \in [-1, 1]:

y=π2y = \dfrac{\pi}{2}


Since y=π2y = \dfrac{\pi}{2} is a constant:

dydx=0\dfrac{dy}{dx} = \boxed{0}

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