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

Answer 1

Isaiah Allen

Knowing the chain rule for derivatives, you can see that

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

and

#dy/dt=4t-4#

Thus, #f'(t)=(2te^(t^2-1)-e^(t), 4t-4)#

In this case, the "trick" is to apply the chain rule to the the first term of the expression

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

to obtain the derivative.

The chain rule formula used in this problem comes from

#d/dx[e^(z)]=e^(z) dz/dx#

where

#z=t^2-1#

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Answer 2

Charles Arocha

To find the derivative of the vector-valued function ( \mathbf{f}(t) = (e^{t^2-1} - e^t, 2t^2 - 4t) ), differentiate each component of the vector separately:

Differentiate the first component ( e^{t^2-1} - e^t ) with respect to ( t ): [ \frac{d}{dt}(e^{t^2-1} - e^t) = \left(2te^{t^2-1} - e^t\right) - e^t ]
Differentiate the second component ( 2t^2 - 4t ) with respect to ( t ): [ \frac{d}{dt}(2t^2 - 4t) = 4t - 4 ]

Combine the derivatives of the components to form the derivative of ( \mathbf{f}(t) ): [ \mathbf{f}'(t) = \left(2te^{t^2-1} - 2e^t, 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.