For the first derivative: If you had 1∫xln(t2)dt, the derivative would just be ln(x2). But you don't have just a plain "x" on the upper bound so you must use the chain rule. The derivative of 3x is 3 so the derivative of 1∫3xln(t2)dt is 3ln(x2). Note that you are just taking the derivative of the upper bound when you use the chain rule, not the upper bound times the derivative of what's inside the "main function".
So, now, use your calculator to find the derivative of 3ln(x2) at x = 1 and you get 6