source : socratic.org

## How do you find the derivative of G(x)=int (e^(t^2))dt from [1,x]?

You should get #e^(x^2)#.

You can use Leibniz’s integral rule.

#(dG(x))/(dx) -= d/(dx) [int_(g(x))^(h(x)) f(x,t)dt]#

#= int_(g(x))^(h(x)) (delf)/(delx)dt + f(x,h(x))(dh)/(dx) – f(x,g(x))(dg)/(dx)#

In this case, #g(x) = 1# and #h(x) = x#, so:

#(dg)/(dx) = 0#

#(dh)/(dx) = 1#

#f(x,h(x)) = f(x,t = x) = e^(x^2)#

#f(x,g(x)) = f(x,t = 1) = e^(1^2) = e#

#(delf)/(delx) = (del)/(delx)[e^(t^2)]_x = 0#

Therefore:

#color(blue)((dG)/(dx)) = cancel(int_(1)^(x) 0dt)^(0) + e^(x^2)cdot1 – cancel(ecdot0)#

#= color(blue)(e^(x^2))#

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