$\dfrac{d}{dx}(\sin^{-1} x)$ equals:
A$\dfrac{-1}{\sqrt{1-x^2}}$
B$\dfrac{1}{1+x^2}$
C$\dfrac{1}{\sqrt{1+x^2}}$
D$\dfrac{1}{\sqrt{1-x^2}}$
Answer & Solution
Correct answer: D. $\dfrac{1}{\sqrt{1-x^2}}$
If $y = \sin^{-1}x$, then $\sin y = x$, so $\cos y \cdot dy/dx = 1$, $dy/dx = 1/\cos y = 1/\sqrt{1 - x^2}$ (taking principal value).
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