Find \( \frac{d y}{d t} \) \[ y=2 t\left(3 t^{3}-7\right)^{5} \] \( \frac{d y}{\frac{d t}{x}}=\square \)

We start with \[ y = 2t\left(3t^3 - 7\right)^5. \] Using the product rule, let \[ u = 2t \quad \text{and} \quad v = \left(3t^3 - 7\right)^5. \] Then \[ \frac{dy}{dt} = u' \cdot v + u \cdot v'. \] 1. Differentiate \( u \): \[ u' = \frac{d}{dt}(2t) = 2. \] 2. Differentiate \( v \) using the chain rule: \[ v = \left(3t^3 - 7\right)^5, \] so \[ v' = 5\left(3t^3 - 7\right)^4 \cdot \frac{d}{dt}(3t^3 - 7). \] Now, compute \[ \frac{d}{dt}(3t^3 - 7)= 9t^2. \] Thus, \[ v' = 5\left(3t^3 - 7\right)^4 \cdot 9t^2 = 45t^2\left(3t^3 - 7\right)^4. \] 3. Combine the derivatives: \[ \frac{dy}{dt} = 2\left(3t^3 - 7\right)^5 + 2t \cdot 45t^2\left(3t^3 - 7\right)^4. \] Simplify the second term: \[ 2t \cdot 45t^2 = 90t^3. \] So, \[ \frac{dy}{dt} = 2\left(3t^3 - 7\right)^5 + 90t^3\left(3t^3 - 7\right)^4. \] 4. Factor out the common term \( 2\left(3t^3 - 7\right)^4 \): \[ \frac{dy}{dt} = 2\left(3t^3 - 7\right)^4\left[\left(3t^3 - 7\right) + 45t^3\right]. \] Simplify inside the brackets: \[ \left(3t^3 - 7\right) + 45t^3 = 48t^3 - 7. \] Thus, the derivative is: \[ \frac{dy}{dt} = 2\left(3t^3 - 7\right)^4\left(48t^3 - 7\right). \]