1) $ y = \cos^{-1}(-5x^3) $
$ \frac{dy}{dx} = \frac{-1}{\sqrt{1 - (-5x^3)^2}} \cdot (-15x^2) = \frac{15x^2}{\sqrt{1 - 25x^6}} $
2) $ y = \sin^{-1}(-2x^2) $
$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (-2x^2)^2}} \cdot (-4x) = \frac{-4x}{\sqrt{1 - 4x^4}} $
3) $ y = \tan^{-1}(2x^4) $
$ \frac{dy}{dx} = \frac{1}{1 + (2x^4)^2} \cdot (8x^3) = \frac{8x^3}{1 + 4x^8} $
4) $ y = \csc^{-1}(4x^2) $
$ \frac{dy}{dx} = \frac{-1}{|4x^2|\sqrt{(4x^2)^2 - 1}} \cdot (8x) = \frac{-8x}{4x^2\sqrt{16x^4 - 1}} = \frac{-2}{x\sqrt{16x^4 - 1}} \quad (x \ne 0) $
5) $ y = (\sin^{-1}(5x^2))^3 $
$ \frac{dy}{dx} = 3(\sin^{-1}(5x^2))^2 \cdot \frac{1}{\sqrt{1 - (5x^2)^2}} \cdot (10x) = \frac{30x(\sin^{-1}(5x^2))^2}{\sqrt{1 - 25x^4}} $
6) $ y = \sin^{-1}((3x^2 + 1)^3) $
$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - ((3x^2 + 1)^3)^2}} \cdot 3(3x^2 + 1)^2 \cdot (6x) = \frac{18x(3x^2 + 1)^2}{\sqrt{1 - (3x^2 + 1)^6}} $
7) $ y = (\cos^{-1}(4x^2))^2 $
$ \frac{dy}{dx} = 2(\cos^{-1}(4x^2)) \cdot \frac{-1}{\sqrt{1 - (4x^2)^2}} \cdot (8x) = \frac{-16x \cos^{-1}(4x^2)}{\sqrt{1 - 16x^4}} $
8) $ y = \cos^{-1}((-2x^3 - 3)^3) $
$ \frac{dy}{dx} = \frac{-1}{\sqrt{1 - ((-2x^3 - 3)^3)^2}} \cdot 3(-2x^3 - 3)^2 \cdot (-6x^2) = \frac{18x^2(-2x^3 - 3)^2}{\sqrt{1 - (-2x^3 - 3)^6}} $
Parent Tip: Review the logic above to help your child master the concept of inverse trigonometric functions worksheet.
