Find $\dfrac{dy}{dx}$ if $y = \dfrac{2x + 1}{x - 1}$.

Question

Accepted Answer

$\dfrac{-3}{(x - 1)^2}$. Use the quotient rule $\dfrac{u'v - uv'}{v^2}$ with $u = 2x + 1$ and $v = x - 1$: numerator $2(x - 1) - (2x + 1)(1) = 2x - 2 - 2x - 1 = -3$, so $\dfrac{dy}{dx} = \dfrac{-3}{(x-1)^2}$. Reversing the order of the two terms in the numerator gives $+3$.

Answer

A) $\dfrac{3}{(x - 1)^2}$

Answer

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

Answer

C) $\dfrac{4x - 1}{(x - 1)^2}$

Answer

D) $\dfrac{-3}{(x - 1)^2}$

Find $\dfrac{dy}{dx}$ if $y = \dfrac{2x + 1}{x - 1}$.

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