2022-01-14

f(t)=sin(2t)-t cos(t)

Vasquez

Find the derivative of the following via implicit differentiation:
$\frac{d}{dt}\left(f\left(t\right)\right)=\frac{d}{dt}\left(-t\mathrm{cos}\left(t\right)+\mathrm{sin}\left(2t\right)\right)$

Using the chain rule, $\frac{d}{dt}\left(f\left(t\right)\right)=\frac{df\left(u\right)}{du}\frac{du}{dt}$, where $u=t$ and $\frac{d}{du}\left(f\left(u\right)\right)={f}^{\prime }\left(u\right):$
$\left(\frac{d}{dt}\left(t\right)\right){f}^{\prime }\left(t\right)=\frac{d}{dt}\left(-t\mathrm{cos}\left(t\right)+\mathrm{sin}\left(2t\right)\right)$

The derivative of t is 1:
$1{f}^{\prime }\left(t\right)=\frac{d}{dt}\left(-t\mathrm{cos}\left(t\right)+\mathrm{sin}\left(2t\right)\right)$

Differentiate the sum term by term and factor out constants:
${f}^{\prime }\left(t\right)=\left(-\left(\frac{d}{dt}\left(t\mathrm{cos}\left(t\right)\right)\right)+\frac{d}{dt}\left(\mathrm{sin}\left(2t\right)\right)\right)$

Use the product rule, $\frac{d}{dt}\left(uv\right)=v\frac{du}{dt}+u\frac{dv}{dt}$, where $u=t$ and $v=cos\left(t\right):$
${f}^{\prime }\left(t\right)=\frac{d}{dt}\left(\mathrm{sin}\left(2t\right)\right)-\left(\mathrm{cos}\left(t\right)\left(\frac{d}{dt}\left(t\right)\right)+t\left(\frac{d}{dt}\left(\mathrm{cos}\left(t\right)\right)\right)\right)$

Simplify the expression:
${f}^{\prime }\left(t\right)=-\mathrm{cos}\left(t\right)\left(\frac{d}{dt}\left(t\right)\right)-t\left(\frac{d}{dt}\left(\mathrm{cos}\left(t\right)\right)\right)+\frac{d}{dt}\left(\mathrm{sin}\left(2t\right)\right)$

The derivative of t is 1:

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