Anyone have any idea?

A curve in the (x, y) plane is defined by the equation
y= integral from 0 to g(x) e^((x^2)- (t^2)) where g(0)=1.

If the curve has slope −3 at the point x = 0, find the value of g′(0).

Question

Anyone have any idea?
 
A curve in the (x, y) plane is defined by the equation
y= integral from 0 to g(x) e^((x^2)- (t^2)) where g(0)=1.

If the curve has slope −3 at the point x = 0, find the value of g′(0).

barrycarter · Answer

How far have you gotten? Hint: fundamental theorem of calculus

anonymous · Answer

since there is "g prime y" , i can say this is diffrential calculus, but how about the "integral from 0 to . ...." , kinda confusing. .

anonymous · Answer

I think I'm suppose to find the integral for the function .. and then I dont know what to do.

anonymous · Answer

i remember,  y'=m (the first derivative of y is equal to the slope), therefor if m=-3, then y prime is -3. .

dumbcow · Answer

not sure, i believe g(x) is the upper limit of the integral
so g(x) is not defined

t is treated as a constant unless this is integrated with respect to t ??

anonymous · Answer

sorry... maybe i didn't type it clearly

dumbcow · Answer

i'm stuck because you can't integrate e^t^2 using elementary functions