MCQ
$\frac{d}{d x}\left[\tan ^{-1}(\sin x)\right]=$
  • $\frac{\cos x}{1+\sin ^2 x}$
  • B
    $\frac{\sin x}{1+\sin ^2 x}$
  • C
    $\frac{-\cos x}{1+\sin ^2 x}$
  • D
    $\frac{-\sin x}{1+\sin ^2 x}$

Answer

Correct option: A.
$\frac{\cos x}{1+\sin ^2 x}$
A

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