What is the derivative of $f(t) = (t/(t+1) , 1/(t^2-t) )$ ?

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Answer 1 · 2018-01-20T23:05:06

$(dy)/(dx)=((t+1)^2(-2t+1))/(t^2-t)^2$

We know that $x=t/(t+1)$ and $y=1/(t^2-t)$

$(dy)/(dx)=(dy)/(dt)-:(dx)/(dt)$

$(dy)/(dt)=((t^2-t)d/dt[1]-1d/dt[t^2-t])/(t^2-t)^2$

$color(white)((dy)/(dt))=(0(t^2-t)-1(2t-1))/(t^2-t)^2$

$color(white)((dy)/(dt))=(-2t+1)/(t^2-t)^2$

$(dx)/(dt)=((t+1)d/dt[t]-td/dt[t+1])/(t+1)^2$

$color(white)((dx)/(dt))=(1(t+1)-t(1))/(t+1)^2$

$color(white)((dx)/(dt))=(t+1-t)/(t+1)^2$

$color(white)((dx)/(dt))=1/(t+1)^2$

$(dy)/(dx)=((-2t+1)/(t^2-t)^2)/(1/(t+1)^2)$

$color(white)((dy)/(dx))=((t+1)^2(-2t+1))/(t^2-t)^2$

What is the derivative of f(t) = (t/(t+1) , 1/(t^2-t) ) f(t) = (t/(t+1) , 1/(t^2-t) ) ?