There are two methods to do this.
Recall that #d/dxa^(f(x))# where #a# is a constant and #f(x)# is some function is #a^f(x)*f'(x)*ln(a)#. This is really just the Chain Rule used to differentiate an exponential.
Thus,
#d/dx3^(sinx+cosx)=3^(sinx+cosx)*d/dx(sinx+cosx)*ln(3)=ln(3)3^(sinx+cosx)(cosx-sinx)#
Or, use logarithmic differentiation:
#f(x)=y#
#y=3^(sinx+cosx)#
Let's apply the natural logarithm to both sides:
#lny=ln(3^(sinx+cosx))#
We can rewrite this as
#lny=(sinx+cosx)*ln(3)#
Because #ln(a^b)=bln(a)#.
Let's differentiate both sides with respect to #x:#
#d/dx(lny)=d/dx((ln3)(sinx+cosx))#
#dy/dx * 1/y=(ln3)*(cosx-sinx)#,
as #d/dxsinx=cosx,d/dxcosx=-sinx#
Let's solve for #dy/dx#, meaning we must multiply both sides by #y:#
#dy/dx=y(ln3*(cosx-sinx))#
It's better to have our equation in terms of only #x.# Replace #y# with #3^(sinx+cosx)#.
#dy/dx=3^(sinx+cosx)(ln3*(cosx-sinx))#
#dy/dx=ln(3)3^(sinx+cosx)(cosx-sinx)#
#d/dx3^(sinx+cosx)=ln(3)3^(sinx+cosx)(cosx-sinx)#
So, our two answers from the different methods are really the same.
