How do you find the derivative of #y=( (2x^3)(e^(sinx))(2^(cosx)) ) / (tanx-3^x)#?

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Answer 1 · 2018-07-16T15:30:11

#y'=(u'v-uv')/v^2# where #u'=6x^2e^sin(x)2^cos(x)+2x^3e^sin(x)cos(x)2^cos(x)+2^cos(x)ln(2)(-sin(x))#
#v=tan(x)-3^x#
#v'=1+tan^2(x)-3^xln(3)#

We Need the rule

#(uvw)'=u'vw+uv'w+uvw'#

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

#u'=6x^2e^sin(x)2^cos(x)+2x^3e^sin(x)cos(x)2^cos(x)+2^cos(x)ln(2)*(-sin(x))#

#v'=1+tan^2(x)-3^xln(3)#
Note that

#(tan(x))'=1+tan^2(x)#

How do you find the derivative of #y=( (2x^3)*(e^(sinx))*(2^(cosx)) ) / (tanx-3^x)#?