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