The first derivative of the function $[{cos}^{- 1} (sin x) + x^{2}]$ with respect to $x$ at $x = 0$ is

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Solution

$y = {cos}^{- 1} (sin x) + x^{2}$
$y^{'} = \frac{- 1}{\sqrt{1 - {sin}^{2} x}} \times cos x + 2 x$
Hence,
$y^{'} (0) = \frac{- 1}{\sqrt{1 - 0}} \times 1 + 2 (0) = - 1$

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