Question

find \\( \frac{d y}{d t} \\)
\\( y=(t \sec t)^{11} \\)
\\( \frac{d y}{d t}= \\)

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = u^{n}\), then \(\frac{dy}{dt}=n\cdot u^{n - 1}\cdot\frac{du}{dt}\). Here \(y=(t\sec t)^{11}\), so let \(u = t\sec t\) and \(n = 11\). Then \(\frac{dy}{dt}=11(t\sec t)^{10}\cdot\frac{d}{dt}(t\sec t)\).

Step2: Apply the product rule

The product rule \(\frac{d}{dt}(uv)=u\frac{dv}{dt}+v\frac{du}{dt}\). For \(u = t\) and \(v=\sec t\), \(\frac{d}{dt}(t\sec t)=t\frac{d}{dt}(\sec t)+\sec t\frac{d}{dt}(t)\).
Since \(\frac{d}{dt}(\sec t)=\sec t\tan t\) and \(\frac{d}{dt}(t) = 1\), we have \(\frac{d}{dt}(t\sec t)=t\sec t\tan t+\sec t\).

Step3: Combine the results

Substitute \(\frac{d}{dt}(t\sec t)\) back into the expression from Step1:

$$
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$$

Answer:

\(11\sec t(t\sec t)^{10}(t\tan t + 1)\)