1. You simply substitute x for t in the function (in other words just take the first half of your answer, and get rid of the dt).
2. Almost. There shouldn't be a minus sign, and the "e" disappeared.
well actually i have two quick questions. i am doing a problem and i wanted to make sure my thought process was correct.
1. when you have an integral, lets say inside the integral is 3*e^((1/2)*sin (2t)dt) with the limits running from 0 to x, if I wanted to take the derivative of this integral then that would simply remove the integral and leave me with the function at the endpoints of 0, and x? so i would have 3*e^((1/2)*sin (2(x)) dt - 3*e^((1/2)*sin (2(0)) dt ? is this correct?
2. my second question is on taking the derivative of the function y= e^(-(1/2)*sin(2x)). to do this derivative i could say -2*(1/2)*cos(2x)?
thanks in advance
i am sorry let me retype the original problem again...
y= e^((-1/2)*sin(2x)) * (integral(3*e^((1/2)*sin(2t)) dt+ c1)) i hope this is better. the first part is y = e raised to the stuff, where the stuff is -1/2 times sin 2x, this is times the integral of
stuff, where the stuff equals 3 times e raised to the 1/2 times sin 2t, dt + c1. so to take the derivative of this function, i used the formula that halls posted earlier.
y' = would be just the product rule of the function. and after doing this problem i ordered some books on latex, i will work on a more suitable notation, that is more clearly understood.