We have #d/dxsec^-1(e^(2x))#.
We can apply the chain rule, which states that for a function #f(u)#, its derivative is #(df)/(du)*(du)/dx#.
Here, #f=sec^-1(u)#, and #u=e^(2x)#.
#d/dxsec^-1(u)=1/(sqrt(u^2)sqrt(u^2-1))#. This is a common derivative.
#d/dxe^(2x)#. Chain rule again, here #f=e^u# and #x=2x#. The derivative of #e^u# is #e^u#, and the derivative of #2x# is #2#.
But here, #u=2x#, and so we finally have #2e^(2x)#.
So #d/dxe^(2x)=2e^(2x)#.
Now we have:
#(2e^(2x))/(sqrt(u^2)sqrt(u^2-1))#, but since #u=e^(2x)#, we have:
#(2e^(2x))/(sqrt((e^(2x))^2)sqrt((e^(2x))^2-1))#
#(2e^(2x))/(e^(2x)sqrt((e^(4x))-1))#
#2/(sqrt(e^(4x)-1))#, our derivative.
