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

Answer 1

Ezra Adams

#(1-e^t)/(3t^2-3e^(3t-1))#

This is a parametric function where #x=t^3-e^(3t-1)# and #y=t-e^t#

and #(dy)/(dt)=1-e^t# and #(dx)/(dt)=3t^2-3e^(3t-1)#

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

= #(1-e^t)/(3t^2-3e^(3t-1))#

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Answer 2

Luke Alvarez

To find the derivative of the given function ( f(t) = (t^3 - e^{3t-1}, t - e^t) ), we differentiate each component with respect to ( t ) separately using the rules of differentiation.

For the first component: [ \frac{d}{dt}(t^3 - e^{3t-1}) = 3t^2 - 3e^{3t-1} ]

For the second component: [ \frac{d}{dt}(t - e^t) = 1 - e^t ]

So, the derivative of the given function is: [ f'(t) = \left(3t^2 - 3e^{3t-1}, 1 - e^t\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.