If y=(2x−1)x2+3y = (2x - 1)\sqrt{x^2 + 3}y=(2x−1)x2+3, what is y′y'y′?
4x2−x+6x2+3\displaystyle\frac{4x^2 - x + 6}{\sqrt{x^2 + 3}}x2+34x2−x+6
xx2+3\displaystyle\frac{x}{\sqrt{x^2 + 3}}x2+3x
2x2+32\sqrt{x^2 + 3}2x2+3
2x2−x+3x2+3\displaystyle\frac{2x^2 - x + 3}{\sqrt{x^2 + 3}}x2+32x2−x+3
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AI Tutor 3:59 AM
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