To find the derivative of this expression, we'll use the product rule, which states:
#d/dx(uv)=u'v+uv'#
Where #u# and #v# are functions of some variable (usually #x#).
We can see that the problem satisfies the two basic requirements of the product rule: both #e^x# and #ln(sinx)# are functions of #x#, and they're being multiplied together. So we can let #u=e^x# and #v=ln(sinx)# and apply the rule. First, we'll have to find the derivatives of these two functions:
#d/dx(e^x)=e^x#
#d/dx(ln(sinx))=cosx*1/sinx=cosx/sinx=cotx->#using chain rule and #lnx# rule
Now we have #u# #(e^x)#, #u'# #(e^x)#, #v# #(ln(sinx))#, and #v'# #(cotx)#. All that is left is to plug these guys into the formula and simplify:
#dy/dx=(e^x)(ln(sinx))+(e^x)(cotx)#
#=e^x(ln(sinx)+cotx)#
Remember that #u# and #v# don't have to be functions of #x#; they can be functions of anything as long as it's the same variable (as in #3z^2cosz#).
