If $y = \frac{x^2 + 1}{x - 1}$, find $\frac{dy}{dx}$.

Question

Accepted Answer

$\frac{x^2 - 2x - 1}{(x-1)^2}$. Using the quotient rule with u = x² + 1 and v = x - 1: dy/dx = [(2x)(x-1) - (x²+1)(1)] / (x-1)². The numerator = 2x² - 2x - x² - 1 = x² - 2x - 1. Option B has a sign error giving +2x, while D gives (x-1)² which would mean the numerator factors perfectly — it does not here.

Answer

A) $\frac{x^2 - 2x - 1}{(x-1)^2}$

Answer

B) $\frac{x^2 + 2x - 1}{(x-1)^2}$

Answer

C) $\frac{2x}{1}$

Answer

D) $\frac{x^2 - 2x + 1}{(x-1)^2}$

If $y = \frac{x^2 + 1}{x - 1}$, find $\frac{dy}{dx}$.

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