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

Answer 1

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

The derivative of a parametric function is defined by: #" "# #dy/dx=(dy/dt)/(dx/dt)# #" "# Here #x=t^2-1" "#and#" "y=te^(2t-1)# #" "# #dx/dt=2t# #" "# #dy/dt=e^(2t-1)+2te^(2t-1)=e^(2t-1)(1+2t)# #" "# #dy/dx=(dy/dt)/(dx/dt)=(e^(2t-1)(1+2t))/(2t)#

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Answer 2

Ezekiel Alexander

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

[ f'(t) = \left(2t, e^{2t-1} + 2te^{2t-1}\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.