What is the derivative of $f (t) = (e^{t^{2} - 1} - t, 2 t^{3} - 4 t)$ $f(t) = (e^(t^2-1)-t, 2t^3-4t )$ ?

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Answer 1 · 2017-04-08T21:04:01

( $2 t e^{t^{2} - 1}$ $2te^(t^2-1)$ , $6 t^{2} - 4$ $6t^(2)-4$ )

To find the derivative of a parametric you just find the derivative of each component separately.

Generally for something in the form $e^{u}$ $e^u$ the derivative is $u'e^u$
$e^(t^2-1)$ derivative is $2te^(t^2-1)$

For $2t^3-4t$ the derivative is in a log way
$3*2t^(3-1)-4t^(1-1)$
$6t^(2)-4$

All together it becomes:
( $x',y'$ )
( $2te^(t^2-1)$ , $6t^(2)-4$ )

Answer 2 · 2017-04-09T14:12:30

$(dy)/(dx)=(6t^2-4)/(2te^(t^2-1)-1)$

This is an equation given in parametric form, where

$x=e^(t^2-1)-t$ and $y=2t^3-4t$

In such cases $(dy)/(dx)=((dy)/(dt))/((dx)/(dt))$

Here $(dx)/(dt)=e^(t^2-1)xx2t-1=2te^(t^2-1)-1$

and $(dy)/(dt)=6t^2-4$

Hence $(dy)/(dx)=(6t^2-4)/(2te^(t^2-1)-1)$

What is the derivative of f(t) = (e^(t^2-1)-t, 2t^3-4t ) f(t)=(et2−1−t,2t3−4t)f(t) = (e^(t^2-1)-t, 2t^3-4t ) ?