4t³sint

Tomas Dela cruz 2022-05-31 Answered

4t³sint

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Answers (1)

karton
Answered 2022-07-07 Author has 439 answers

Since 4 is constant with respect to t, the derivative of 4t3sin(t) with respect to t is 4ddt[t3sin(t)].

Differentiate using the Product Rule which states that ddt[f(t)g(t)] is f(t)ddt[g(t)]+g(t)ddt[f(t)] 

where f(t)=t3 and g(t)=sin(t).

4(t3ddt[sin(t)]+sin(t)ddt[t3])

The derivative of sin(t) with respect to t is cos(t).

4(t3cos(t)+sin(t)ddt[t3])

Differentiate using the Power Rule which states that ddt[tn] is ntn-1 where n=3.

4(t3cos(t)+sin(t)(3t2))

Simplify.

4t3cos(t)+12t2sin(t)

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