find the derivative of the function.\ns = sinleft(\frac{17pi t}{6}\right)+cosleft(\frac{17pi…

Explanation:

Step1: Apply sum - rule of derivatives

$\frac{ds}{dt}=\frac{d}{dt}\sin\left(\frac{17\pi t}{6}\right)+\frac{d}{dt}\cos\left(\frac{17\pi t}{6}\right)$

Step2: Apply chain - rule for $\frac{d}{dt}\sin\left(\frac{17\pi t}{6}\right)$

Let $u = \frac{17\pi t}{6}$, then $\frac{d}{dt}\sin\left(\frac{17\pi t}{6}\right)=\cos(u)\cdot\frac{du}{dt}$. Since $\frac{du}{dt}=\frac{17\pi}{6}$, we have $\frac{d}{dt}\sin\left(\frac{17\pi t}{6}\right)=\frac{17\pi}{6}\cos\left(\frac{17\pi t}{6}\right)$

Step3: Apply chain - rule for $\frac{d}{dt}\cos\left(\frac{17\pi t}{6}\right)$

Let $u=\frac{17\pi t}{6}$, then $\frac{d}{dt}\cos\left(\frac{17\pi t}{6}\right)=-\sin(u)\cdot\frac{du}{dt}$. Since $\frac{du}{dt}=\frac{17\pi}{6}$, we have $\frac{d}{dt}\cos\left(\frac{17\pi t}{6}\right)=-\frac{17\pi}{6}\sin\left(\frac{17\pi t}{6}\right)$

Step4: Combine the results

$\frac{ds}{dt}=\frac{17\pi}{6}\cos\left(\frac{17\pi t}{6}\right)-\frac{17\pi}{6}\sin\left(\frac{17\pi t}{6}\right)=\frac{17\pi}{6}\left(\cos\left(\frac{17\pi t}{6}\right)-\sin\left(\frac{17\pi t}{6}\right)\right)$

Answer:

$\frac{17\pi}{6}\left(\cos\left(\frac{17\pi t}{6}\right)-\sin\left(\frac{17\pi t}{6}\right)\right)$