Questions: Find dy / dt. y=t^4(t^6+8)^6

Solution Steps

To find \( \frac{dy}{dt} \) for the given function \( y = t^4 (t^6 + 8)^6 \), we will use the product rule and the chain rule of differentiation. The product rule states that \( \frac{d}{dt}[u \cdot v] = u' \cdot v + u \cdot v' \). Here, \( u = t^4 \) and \( v = (t^6 + 8)^6 \). We will differentiate each part separately and then combine them using the product rule.

Given the function: \[ y = t^4 (t^6 + 8)^6 \]

To find \( \frac{dy}{dt} \), we use the product rule: \[ \frac{d}{dt} [u \cdot v] = u' \cdot v + u \cdot v' \] where \( u = t^4 \) and \( v = (t^6 + 8)^6 \).

First, differentiate \( u \): \[ u = t^4 \implies u' = 4t^3 \] Next, differentiate \( v \) using the chain rule: \[ v = (t^6 + 8)^6 \implies v' = 6(t^6 + 8)^5 \cdot \frac{d}{dt}(t^6 + 8) = 6(t^6 + 8)^5 \cdot 6t^5 = 36t^5(t^6 + 8)^5 \]

Combine the derivatives using the product rule: \[ \frac{dy}{dt} = u' \cdot v + u \cdot v' = 4t^3 (t^6 + 8)^6 + t^4 \cdot 36t^5 (t^6 + 8)^5 \]

Simplify the expression: \[ \frac{dy}{dt} = 4t^3 (t^6 + 8)^6 + 36t^9 (t^6 + 8)^5 \] Factor out the common terms: \[ \frac{dy}{dt} = t^3 (t^6 + 8)^5 (4(t^6 + 8) + 36t^6) = t^3 (t^6 + 8)^5 (4t^6 + 32 + 36t^6) = t^3 (t^6 + 8)^5 (40t^6 + 32) \]

Final Answer