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

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Answer 1 · 2016-06-03T06:31:30

#(dy)/(dx)=1-2e^(2t^2-2)#

This is parametric form of equation. In parametric form of equation,

#(dy)/(dx)=(dy)/(dt)/(dx)/(dt)#

Now as #y=t^2+1-e^(2t^2-2)#

#(dy)/(dt)=2t-(e^(2t^2-2)xx4t)#

= #2t(1-2e^(2t^2-2))#

As #x=t^2-1#

#(dx)/(dt)=2t#

Hence, #(dy)/(dx)=(2t(1-2e^(2t^2-2)))/(2t)#

or #(dy)/(dx)=1-2e^(2t^2-2)#