find the derivative of the given function.\nr = \frac{sinleft(3t-\frac{pi}{2}\right)}{4t}\n\frac{dr}{dt}=squa…

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $r=\frac{u}{v}$, then $\frac{dr}{dt}=\frac{u'v - uv'}{v^{2}}$. Here, $u = \sin(3t-\frac{\pi}{2})$ and $v = 4t$.

Step2: Find $u'$

Using the chain - rule, if $y=\sin(u)$ and $u = 3t-\frac{\pi}{2}$, then $\frac{dy}{du}=\cos(u)$ and $\frac{du}{dt}=3$. So, $u'=\cos(3t - \frac{\pi}{2})\cdot3=3\cos(3t-\frac{\pi}{2})$.

Step3: Find $v'$

Since $v = 4t$, then $v'=4$.

Step4: Substitute into quotient - rule

$\frac{dr}{dt}=\frac{3\cos(3t-\frac{\pi}{2})\cdot4t-\sin(3t-\frac{\pi}{2})\cdot4}{(4t)^{2}}=\frac{12t\cos(3t - \frac{\pi}{2})-4\sin(3t-\frac{\pi}{2})}{16t^{2}}=\frac{3t\cos(3t-\frac{\pi}{2})-\sin(3t-\frac{\pi}{2})}{4t^{2}}$.

Answer:

$\frac{3t\cos(3t-\frac{\pi}{2})-\sin(3t-\frac{\pi}{2})}{4t^{2}}$