6. $[-/ 1$ Points $]$ DETAILS SCALC9 4.3.017. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Use part one of the fundamental theorem of calculus to find the derivative of the function. \[ y=\int_{4}^{3 x+5} \frac{t}{1+t^{3}} d t \] $y^{\prime}=$ Need Help? Read It Submit Answer

Step 1 ：The problem is asking for the derivative of the function \(y=\int_{4}^{3 x+5} \frac{t}{1+t^{3}} dt\).

Step 2 ：According to the first part of the Fundamental Theorem of Calculus, the derivative of the integral from a constant to a function of \(x\) is just the integrand evaluated at that function of \(x\), multiplied by the derivative of that function of \(x\).

Step 3 ：So, we need to substitute \(3x+5\) into \(\frac{t}{1+t^{3}}\) and then multiply by the derivative of \(3x+5\), which is \(3\).

Step 4 ：The derivative of the function is \(-9*(3x + 5)^3/((3x + 5)^3 + 1)^2 + 3/((3x + 5)^3 + 1)\).

Step 5 ：So, the final answer is \(y^\prime = \boxed{-9*(3x + 5)^3/((3x + 5)^3 + 1)^2 + 3/((3x + 5)^3 + 1)}\).