Alternative approach (basically the same as the one already presented):

Note that #y=arcsin(e^x)# can be manipulated to say that #sin(y)=e^x#.

Take the derivative of both sides with respect to #x#. Recall that the chain rule will be used on the left side: #cos(y)dy/dx=e^x#

Solving for #dy/dx# gives #dy/dx=e^x/cos(y)#. We know that #sin(y)=e^x#, so we can write #cos(y)# as #cos(y)=sqrt(1-sin^2(y)#. Note that this relationship comes from the Pythagorean identity #sin^2(y)+cos^2(y)=1#.

Thus, we have #dy/dx=e^x/sqrt(1-sin^2(y))#. And, since we know that #sin(y)=e^x#, this gives a fully simplified derivative of #dy/dx=e^x/sqrt(1-e^(2x))#.