How do you find the derivative of $f(t) = t^2sin t$ ?

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Answer 1 · 2018-03-02T14:39:50

$=> f'(t) = 2tsint + t^2 cost$

For this problem we must use a rule called the product rule:

$d/(dt) ( h(t) g(t) ) = h'(t)g(t) + h(t)g'(t)$

Where $h'(t) = d/(dt) (h(t) )$ if your not aware

So in this $h(t) = t^2$ and $g(t) = sint$

$=> f'(t) = (d/(dt) (t^2) * sint) + (t^2* d/(dt) ( sint ))$

$=> f'(t) = 2tsint + t^2 cost$

How do you find the derivative of f(t) = t^2sin tf(t) = t^2sin t?