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

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Answer 1 · 2016-01-14T02:55:50

Integrate each part seperately, since they are in a different axis each.

$f'(t)=(2t-cost,-1/(t-1)^2)$

1st part

$(t^2-sint)'=2t-cost$

2nd part

$(1/(t-1))'=((t-1)^-1)'=-1*(t-1)^(-1-1)*(t-1)'=$

$=-(t-1)^(-2)*1=-1/(t-1)^2$

Result

$f'(t)=(2t-cost,-1/(t-1)^2)$

Answer 2 · 2016-01-14T02:57:28

$-1/((2t-cost)(t-1)^2)$

$x(t)=t^2-sint$
$y(t)=1/(t-1)$

$x'(t)=2t-cost$
$y'(t)=-1/(t-1)^2$

To find the derivative of a parametric function, find

$dy/dx=(dy/dt)/(dx/dt)=(y'(t))/(x'(t))=(-1/(t-1)^2)/(2t-cost)=-1/((2t-cost)(t-1)^2)$

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