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

Answer 1

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

Derivative of a parametric function #f(t)=(x(t),y(t))# is given by #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))# As #y(t)=2t^2-4t#, #(dy)/(dt)=4t-4#

and as #x(t)=e^(t^2-1)-t#,, #(dx)/(dt)=e^(t^2-1)xx2t-1=2te^(t^2-1)-1#

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

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Answer 2

Emma Aldridge

The derivative of ( f(t) = (e^{t^2-1} - t, 2t^2 - 4t) ) with respect to ( t ) is:

[ f'(t) = \left( (2t)e^{t^2-1} - 1, 4t - 4 \right) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.