What is the derivative of #f(t) = (t/(t+1) , te^(2t1) ) #?
Hence derivative of
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To find the derivative of the vectorvalued function ( \mathbf{f}(t) = \left( \frac{t}{t+1}, , t e^{2t1} \right) ), we differentiate each component function with respect to ( t ).

Differentiate the first component function ( \frac{t}{t+1} ) with respect to ( t ): [ \frac{d}{dt}\left( \frac{t}{t+1} \right) = \frac{(t+1)(1)  t(1)}{(t+1)^2} = \frac{1}{(t+1)^2} ]

Differentiate the second component function ( t e^{2t1} ) with respect to ( t ) using the product rule: [ \frac{d}{dt}\left( t e^{2t1} \right) = e^{2t1} + t \cdot \frac{d}{dt}(e^{2t1}) ] Using the chain rule ( \frac{d}{dt}(e^{2t1}) = e^{2t1} \cdot \frac{d}{dt}(2t1) = 2e^{2t1} ): [ \frac{d}{dt}\left( t e^{2t1} \right) = e^{2t1} + 2te^{2t1} ]
Therefore, the derivative of ( \mathbf{f}(t) ) is: [ \mathbf{f}'(t) = \left( \frac{1}{(t+1)^2}, , e^{2t1} + 2te^{2t1} \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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