# Differentiating using multiple rules - spot my mistake

I'm trying to figure out my mistake here.

Find the Derivative of $$\frac {x^2}{x-1}$$ Now I want to simplify this and use the product rule instead of the quotient rule: $$(x^2)(x-1)^{-1}$$ Use the product rule: $$[\frac {d}{dx}x^2](x-1)^{-1}+[{\frac {d}{dx}(x-1)^{-1}]}(x^2)$$ Simplify left side: $$\frac {2x}{x-1}$$ Simplify right side by using chain rule: $$-(x-1)^{-2}(x^2)$$ $$-\frac {x^2}{(x-1)^2}$$ Add them all together, result:$$\frac {2x-x^2}{(x-1)-(x-1)^2}$$

The correct answer would have been: $$\frac {x^2-2x}{(x-1)^2}$$

I'm pretty sure, would I've used the quotient rule, the result would've been the correct one. But why did this result in a wrong answer? Thanks in advance for any clarity on the subject.

• Why did you add the denominators? – user328442 Apr 18 '18 at 21:40
• You added the fractions wrong! – B. Mehta Apr 18 '18 at 21:41
• You need to get a common denominator – Prince M Apr 18 '18 at 22:08

$$\frac {2x}{x-1}-\frac {x^2}{(x-1)^2}=\frac {2x(x-1)-x^2}{(x-1)^2}=\frac {x^2-2x}{(x-1)^2}$$
This line is not correct $$\frac {2x-x^2}{(x-1)-(x-1)^2}$$ It should be $$\frac {2x}{(x-1)}-\frac {x^2}{(x-1)^2}=\frac {2x(x-1)-x^2}{(x-1)^2}=\frac {x^2-2x}{(x-1)^2}$$