##y=ln(u)##
##:. dy/(du)=1/u=1/(e^(2x))##
##u=e^(2x)##, ##:. (du)/(dx)=2e^(2x)##
Now, using the …
##(dy)/(du)*(du)/(dx)=1/(e^(2x))*2e^(2x)=2##
##:. (dy)/(dx)=2##
This all worked because:
When ##y=e^(f(x))##
##(dy)/(dx)=f'(x)e^(f(x))##
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