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

Answer 1

Emily Ammerman

#dy/dx=(-3t^2+1)/(2te^(t^2-1)+3)#

#x(t)=e^(t^2-1)+3t# #y(t)=-t^3+t#

#x'(t)=2te^(t^2-1)+3# #y'(t)=-3t^2+1#

The derivative of a parametric function can be found through

#dy/dx=(dy/dt)/(dx/dt)=(y'(t))/(x'(t))=(-3t^2+1)/(2te^(t^2-1)+3)#

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Answer 2

Ezra Adams

To find the derivative of ( f(t) = (e^{t^2-1} + 3t, -t^3 + t) ), differentiate each component function with respect to ( t ):

Differentiate the first component function: ( \frac{d}{dt} (e^{t^2-1} + 3t) = e^{t^2-1} \cdot 2t + 3 )
Differentiate the second component function: ( \frac{d}{dt} (-t^3 + t) = -3t^2 + 1 )

Therefore, the derivative of ( f(t) ) with respect to ( t ) is: ( f'(t) = \left( e^{t^2-1} \cdot 2t + 3, -3t^2 + 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.