What is the derivative of t2 sin (3t - 5) with respect to t?

To find the derivative of the function t^2 sin(3t-5), we will use the product rule of differentiation. The product rule states that the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. So, applying the product rule to the given function, we get: d/dt [t^2 sin(3t-5)] = t^2 d/dt[sin(3t-5)] + sin(3t-5) d/dt[t^2] Now, we need to find the derivatives of sin(3t-5) and t^2. The derivative of sin(3t-5) is cos(3t-5) times the derivative of the inside function (3t-5), which is simply 3. So, d/dt[sin(3t-5)] = 3cos(3t-5). The derivative of t^2 is 2t. So, d/dt[t^2] = 2t. Now, we can substitute these values back into the original equation: d/dt [t^2 sin(3t-5)] = t^2 * 3cos(3t-5) + sin(3t-5) * 2t Simplifying this expression, we get: d/dt [t^2 sin(3t-5)] = 3t^2 cos(3t-5) + 2t sin(3t-5) Therefore, the correct option is: 2t sin (3t - 5) + 3t^2 cos (3t - 5).