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How do you differentiate #t^2sint#?

Answer 1

Christopher Allers

I found: #t[2sin(t)+tcos(t)]#

I would use the Product Rule to get: #f'(t)=2tsin(t)+t^2cos(t)=t[2sin(t)+tcos(t)]#

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Answer 2

Axel Alvarez

To differentiate ( t^2 \sin(t) ), you would use the product rule of differentiation, which states that if you have two functions ( u(t) ) and ( v(t) ), then the derivative of their product ( u(t)v(t) ) with respect to ( t ) is given by ( u'(t)v(t) + u(t)v'(t) ).

Applying the product rule to ( t^2 \sin(t) ):

Let ( u(t) = t^2 ) and ( v(t) = \sin(t) ).

Then, ( u'(t) = 2t ) (derivative of ( t^2 ) with respect to ( t )) and ( v'(t) = \cos(t) ) (derivative of ( \sin(t) ) with respect to ( t )).

Now, applying the product rule:

( (t^2 \sin(t))' = u'(t)v(t) + u(t)v'(t) )

( = (2t)(\sin(t)) + (t^2)(\cos(t)) )

So, the derivative of ( t^2 \sin(t) ) with respect to ( t ) is ( 2t \sin(t) + t^2 \cos(t) ).

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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.