Question

find \\( \frac { d y } { d t } \\).

\\( y = ( t \sin t ) ^ { 9 } \\)

\\( \frac { d y } { d t } = \\)

Explanation:

Step1: Apply the chain rule

Let \(u = t\sin t\), then \(y = u^{9}\). By the chain rule \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\).
First, \(\frac{dy}{du} = 9u^{8}\).

Step2: Differentiate \(u=t\sin t\) using the product rule

The product rule states that if \(u = vw\) (\(v = t\), \(w=\sin t\)), then \(\frac{du}{dt}=v'\cdot w + v\cdot w'\).
Since \(v'=\frac{d}{dt}(t) = 1\) and \(w'=\frac{d}{dt}(\sin t)=\cos t\), we have \(\frac{du}{dt}=\sin t+t\cos t\).

Step3: Substitute \(u\) and \(\frac{du}{dt}\) back into the chain - rule formula

\(\frac{dy}{dt}=9(t\sin t)^{8}(\sin t + t\cos t)\)

Answer:

\(9(t\sin t)^{8}(\sin t + t\cos t)\)