Question

If $\sqrt{1-x^{2} } + \sqrt{1- y^{2}} =x -y $, then $\frac{dy}{dx} = $

• A. $ -\sqrt{\frac{1-y^{2}}{1- x^{2}}} $
• B. $ \sqrt{\frac{1- x^{2}}{1- y^{2}}} $
• C. $ -\sqrt{\frac{1-x^{2}}{1- y^{2}}} $
• D. $ \sqrt{\frac{1-y^{2}}{1- x^{2}}} $

Answer 1

The Correct Option is D

Put \(x = \sin \theta , y = \sin \theta ,\) we get \(\sqrt{1- \sin^{2} \theta } + \sqrt{1 -\sin^{2} \phi } =\sin \theta -\sin \phi\)
\(\Rightarrow \cos\theta +\cos \phi=\sin\theta-\sin\phi\)
\(\Rightarrow 2\cos \left(\frac{\theta+\phi}{2}\right) \cos \left(\frac{\theta-\phi}{2}\right) = 2 \cos \left(\frac{\theta +\phi }{2}\right) \sin \left(\frac{\theta -\phi }{2}\right)\)
\(\Rightarrow \tan\left(\frac{\theta-\phi}{2}\right) =1\Rightarrow \frac{\theta-\phi}{2} =\tan^{-1} \left(1\right) = \frac{\pi}{4}\)
\(\Rightarrow\theta-\phi=\frac{\pi}{2}\)
\(\Rightarrow \sin^{-1} x -\sin^{-1} y = \frac{\pi}{2} \, \, \, \, \, \, \,\) ...(i) 
Differentiating (i) w.r.t. 'x', we get 
\(\frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-y^{2}} } \frac{dy}{dx} =0 \Rightarrow \frac{dy}{dx} = \frac{\sqrt{1-y^{2}}}{\sqrt{1 -x^{2}}}\)

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In terms of geometry, the slope of the tangent line to the function at a particular location is represented by the derivative. 

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The pace at which a function changes in response to an independent variable is known as its derivative. They are characterised as the rate at which one quantity y changes in relation to another, x. 

The primary objects of differentiation are derivatives. 

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