Solving the Differential Equation: $\frac{d}{dt} \frac{t}{(t-1)^2}$

[QuarkCharmer]

Homework Statement

d/dt t/(t-1)^2

Homework Equations

No Chain Rule.

The Attempt at a Solution

\frac{d}{dt} \frac{t}{(t-1)^2}
\frac{t\frac{d}{dt}(t-1)^2 - (t-1)^2 \frac{d}{dt}t}{(t-1)^4}
\frac{t\frac{d}{dt}(t^2-2t+1) - 1(t-1)^2}{(t-1)^4}
\frac{t(2t-2)-(t-1)^2}{(t-1)^4}
\frac{2t^2-2t-t^2+2t-1}{(t-1)^4}
\frac{(t+1)(t-1)}{(t-1)(t-1)^3}
\frac{(t+1)}{(t-1)^3}

The book shows the solution being negative. I can't figure out where I am going wrong here.

 

[ckutlu]

I think you have a problem at the 2nd line of your attempt. You should first take the derivative of the above term when taking the derivatives of fractions.

Should be like this:
\frac{(t-1)^2 \frac{d}{dt}t- t\frac{d}{dt}(t-1)^2}{(t-1)^4}

 

[QuarkCharmer]

Ahh, I see. I'm doing the quotient rule wrong then.

Thanks!