4t³sint

Tomas Dela cruz
2022-05-31
Answered

4t³sint

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karton

Answered 2022-07-07
Author has **439** answers

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

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

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

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

The derivative of $\mathrm{sin}\left(t\right)$ with respect to $t$ is $\mathrm{cos}\left(t\right)$.

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

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

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

Simplify.

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

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