Mathematics · Calculus

If y = √log(x2 + 1) + √log(x2 + 1) + √log(x2 + 1) + ··· where |x| < 1, then dydx is equal to:

Verified Solution4 OptionsUpdated: October 2026👁️1 View

12x(x2 + 1)(2y - 1)✓
21(x2 + 1)(2y - 1)
32x2y - 1
4x2 + 12y - 1

Correct answer

2x(x2 + 1)(2y - 1)

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If y = √log(x2 + 1) + √log(x2 + 1) + √log(x2 + 1) + ··· where |x| < 1, then dydx is equal to:

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If y = √log(x2 + 1) + √log(x2 + 1) + √log(x2 + 1) + ··· where |x| < 1, then dydx is equal to:

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