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How do you find the derivative of #f(t) = t² sin t#?

Answer 1

Cameron Alexander

#f'(t) =t^2cost+2tsint#

#f(t) = t^2sint#

The Product rule states that:

If #f(t) = g(t)h(t)# then #f'(t) = g(t)h'(t)+ g'(t)h(t)#

In this example: #g(t) = t^2# and #h(t) = sint#

Hence, #f'(t) = t^2*d/dtsint + d/dt t^2 * sint#

#= t^2cost+2tsint#

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Answer 2

Charles Arocha

To find the derivative of ( f(t) = t^2 \sin(t) ), you can use the product rule of differentiation, which states that if you have a function ( u(t) ) and a function ( v(t) ), then the derivative of their product ( u(t)v(t) ) is given by ( u'(t)v(t) + u(t)v'(t) ).

Applying the product rule to ( f(t) = t^2 \sin(t) ), where ( u(t) = t^2 ) and ( v(t) = \sin(t) ), the derivative ( f'(t) ) is:

[ f'(t) = (2t \cdot \sin(t)) + (t^2 \cdot \cos(t)) ]

Therefore, the derivative of ( f(t) = t^2 \sin(t) ) is ( f'(t) = 2t \cdot \sin(t) + t^2 \cdot \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.