If y = e log ( sin - 1 x ) + e log ( cos - 1 x ) , 0 x 1 , then:

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Solution

Q.
If $y = e^{\log (\sin^{- 1} x)} + e^{\log (\cos^{- 1} x)}, 0 < x < 1,$ then:

1 $\frac{d y}{d x} = 0$

2 $\frac{d y}{d x} = \frac{π}{2}$

3 $\frac{d y}{d x} = \frac{π}{3}$

4 does not exist

Ans.
(1)

Ans.    $\frac{d y}{d x} = 0$

Explanation:

   $y = e^{\log (\sin^{- 1} x)} + e^{\log (\cos^{- 1} x)}$

$y = \sin^{- 1} x + \cos^{- 1} x$

   $y = 1$

$\frac{d y}{d x} = 0$

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