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Mathematics Question on limits and derivatives

If u=sin⁡(mcos⁡−1x),v=cos⁡(msin⁡−1x)u = \sin (m \cos^{-1} x), v = \cos (m \sin^{-1} x), then dudv=\frac{du}{dv} =

A

1−u21−v2\sqrt{\frac{ 1- u^2}{ 1- v^2}}

B

1−v21−u2\sqrt{\frac{ 1- v^2}{ 1- u^2}}

C

1+u21+v2\sqrt{\frac{ 1 + u^2}{ 1 + v^2}}

D

none of these.

Answer

1−u21−v2\sqrt{\frac{ 1- u^2}{ 1- v^2}}

Explanation

Solution

u=sin⁡(mcos⁡−1x)u =\sin\left(m \cos^{-1}x\right) dudx=cos⁡(mcos⁡−1x).−m1−x2\frac{du}{dx} =\cos\left(m \cos^{-1}x\right) . \frac{-m}{\sqrt{1-x^{2}}} v=cos⁡(msin⁡−1x)v = \cos\left(m\sin^{-1}x\right) dvdx=−sin⁡(msin⁡−1x).m1−x2\frac{dv}{dx} =-\sin \left(m\sin^{-1} x\right). \frac{m}{\sqrt{1-x^{2}}} ∴dudv=du/dxdv/dx=cos⁡(mcos⁡−1x)sin⁡(msin⁡−1x)\therefore \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\cos\left(m \cos^{-1}x\right)}{\sin\left(m \sin^{-1}x\right)} =1−sin⁡2(mcos⁡−1x)1−cos⁡2(msin⁡−1x)=1−u21−v2= \frac{\sqrt{1-\sin^{2}\left(m \cos^{-1}x\right)}}{\sqrt{1-\cos^{2}\left(m\sin^{-1}x\right)}} = \sqrt{\frac{1-u^{2}}{1-v^{2}}}