\(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac uv\right)=\dfrac{v\dfrac{\mathrm{d}u}{\mathrm{d}x}-u\dfrac{\mathrm{d}v}{\mathrm{d}x}}{v^2}\qquad v\ne0\)
Keep the subtraction order when applying the quotient rule.
Example
Differentiate \(y=\dfrac{x^2+1}{3x-1}\) with respect to \(x\), where \(x\ne\dfrac13\).
\(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)\({}=\dfrac{(3x-1)(2x)-(x^2+1)(3)}{(3x-1)^2}\)\({}=\dfrac{6x^2-2x-3x^2-3}{(3x-1)^2}\)\({}=\dfrac{3x^2-2x-3}{(3x-1)^2}\)
Why the rule works
Write \(\dfrac uv=uv^{-1}\). The product and chain rules give \(\dfrac{\mathrm du}{\mathrm dx}v^{-1}-uv^{-2}\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{v\dfrac{\mathrm du}{\mathrm dx}-u\dfrac{\mathrm dv}{\mathrm dx}}{v^2}\).
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