How do you find the derivative of #(e^(2x))/(4^x)#?

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Answer 1 · 2018-06-16T10:36:57

#(e^(2x)(1-ln(2)))/(2^(2x-1))#

#(u/v)'=(u'v-uv')/v^2#

we get

#f'(x)=(e^(2x)2*4^x-e^(2x)*4^x*ln(4))/(4^x)^2#
simplifying we get

#(e^(2x)(1-ln(2)))/(2^(4x-2x-1))#
and this is equal to

#(e^(2x)(1-ln(2)))/(2^(2x-1))#