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)

4t³sint

Tomas Dela cruz 2022-05-31 Answered

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karton

Answered 2022-07-07 Author has 439 answers

Since $4$ is constant with respect to $t$ , the derivative of $4 t^{3} \sin (t)$ with respect to $t$ is $4 \frac{d}{d t} [t^{3} \sin (t)]$ .

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$

where $f (t) = t^{3}$ and $g (t) = \sin (t)$ .

$4 (t^{3} \frac{d}{d t} [\sin (t)] + \sin (t) \frac{d}{d t} [t^{3}])$

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

$4 (t^{3} \cos (t) + \sin (t) \frac{d}{d t} [t^{3}])$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 3$ .

$4 (t^{3} \cos (t) + \sin (t) (3 t^{2}))$

Simplify.

$4 t^{3} \cos (t) + 12 t^{2} \sin (t)$

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