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

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Answer 1 · 2017-04-08T21:04:01

(#2te^(t^2-1)#, #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# 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)#