\(\displaystyle \frac{d}{dx} Sin^{-1} x = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx} Cos^{ -1} x = -\frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx} Tan^{-1} x = \frac{1}{1 + x^2}\)
\(\displaystyle \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}\)
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