What is the derivative of #f(t) = (te^t +e^t, e^tcost ) #?
So:
And
So:
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To find the derivative of the vectorvalued function ( \mathbf{f}(t) = (te^t + e^t, e^t  \cos(t)) ), we differentiate each component of the function separately with respect to ( t ).

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

For the second component: [ \frac{d}{dt}(e^t  \cos(t)) = \frac{d}{dt}(e^t)  \frac{d}{dt}(\cos(t)) ] [ = e^t + \sin(t) ]
So, the derivative of ( \mathbf{f}(t) ) is: [ \mathbf{f}'(t) = \left( (1 + t)e^t + e^t, \ e^t + \sin(t) \right) ]
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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