How do you differentiate #y=3^sinx#?

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Answer 1 · 2016-09-24T17:05:28

#y = 3^sinx#

#lny = ln(3^(sinx))#

#lny = sinxln3#

#1/y(dy/dx) = (cosx xx ln3 + 0 xx sinx)#

#dy/dx = (ln3cosx)/(1/y)#

#dy/dx = y xx ln3cosx#

#dy/dx = 3^sinx xx ln3cosx#

#dy/dx = 3^sinxln3cosx#

Hopefully this helps!