Question

find \\( \frac { d y } { d t } \\).
\\( y = \sin ( \tan ( 7 t - 6 ) ) \\)
\\( \frac { d y } { d t } = \\)

Explanation:

Step1: Apply the chain rule

Let \(u = \tan(7t - 6)\), then \(y=\sin(u)\). By the chain rule \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\).
First, \(\frac{dy}{du}=\cos(u)\) (since the derivative of \(y = \sin(u)\) with respect to \(u\) is \(\cos(u)\)).

Step2: Find \(\frac{du}{dt}\)

Now, \(u=\tan(7t - 6)\). Let \(v = 7t-6\), then \(u = \tan(v)\). By the chain rule \(\frac{du}{dt}=\frac{du}{dv}\cdot\frac{dv}{dt}\).
The derivative of \(u=\tan(v)\) with respect to \(v\) is \(\sec^{2}(v)\), and the derivative of \(v = 7t - 6\) with respect to \(t\) is \(7\). So \(\frac{du}{dt}=7\sec^{2}(7t - 6)\).

Step3: Substitute back

Since \(u=\tan(7t - 6)\), \(\frac{dy}{dt}=\cos(\tan(7t - 6))\cdot7\sec^{2}(7t - 6)\)

Answer:

\(7\sec^{2}(7t - 6)\cos(\tan(7t - 6))\)