If y=(3x−2)x2−1y = (3x - 2)\sqrt{x^2 - 1}y=(3x−2)x2−1, then y′=y' = y′=
6x−2x2−1\displaystyle\frac{6x - 2}{\sqrt{x^2 - 1}}x2−16x−2
3x2−13\sqrt{x^2 - 1}3x2−1
6x2−2x−3x2−1\displaystyle\frac{6x^2 - 2x - 3}{\sqrt{x^2 - 1}}x2−16x2−2x−3
3x2−2xx2−1\displaystyle\frac{3x^2 - 2x}{\sqrt{x^2 - 1}}x2−13x2−2x
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AI Tutor 5:03 AM
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