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

Answer 1

#f'(t)=(-3/(3t+2)^2, -(t^2+1)/(t^2-1)^2)#

Take the derivative component-by-component:

#f'(t)=(d/dt(1/(3t+2)),d/dt(t/(t^2-1)))#

#d/dt(1/(3t+2))=-3/(3t+2)^2# by the chain rule #d/dt(t/(t^2-1))=((t^2-1)(1)-t(2t))/(t^2-1)^2# #=(-t^2-1)/(t^2-1)^2=-(t^2+1)/(t^2-1)^2#

So,

#f'(t)=(-3/(3t+2)^2, -(t^2+1)/(t^2-1)^2)#

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Answer 2

Emma Aldridge

To find the derivative of the given function ( f(t) = \left(\frac{1}{3t+2}, \frac{t}{t^2-1}\right) ), we differentiate each component separately using the rules of calculus.

The derivative of the first component with respect to ( t ) is:

[ \frac{d}{dt}\left(\frac{1}{3t+2}\right) = -\frac{3}{(3t+2)^2} ]

The derivative of the second component with respect to ( t ) is:

[ \frac{d}{dt}\left(\frac{t}{t^2-1}\right) = \frac{(t^2-1) - t(2t)}{(t^2-1)^2} = \frac{-t^2 - 1}{(t^2-1)^2} ]

Therefore, the derivative of the function ( f(t) = \left(\frac{1}{3t+2}, \frac{t}{t^2-1}\right) ) with respect to ( t ) is:

[ f'(t) = \left(-\frac{3}{(3t+2)^2}, \frac{-t^2 - 1}{(t^2-1)^2}\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.