First, we take #f'(x)# and input #pi/12#. If the answer is more than zero, it is increasing, and if it is less than zero, it is decreasing.

We must find #d/dx(4sin(4x-(3pi)/4))#

According to the product rule, #(f*g)'=f'g+fg'#

Here, #f=4# and #g=sin(4x-(3pi)/4)#

But since #d/dx4=0#, the product rule reduces to:

#fg'#, or:

#4*d/dxsin(4x-(3pi)/4)#

According to the chain rule, #(df)/dx=(df)/(du)*(du)/dx#, where #u# is a function within #f#. Here:

#d/dxsin(4x-(3pi)/4)#

#=d/dxsin(u)*d/dx(4x-(3pi)/4)#

#=cos(u)*4#

And as #u=(4x-(3pi)/4)#, we have:

#4cos(4x-(3pi)/4)#

Applying it into #fg'#:

#4*4cos(4x-(3pi)/4)#

#16cos(4x-(3pi)/4)# is our derivative. Inputting #pi/12# for #x#:

#16cos(4*pi/12-(3pi)/4)#

#16cos(pi/3-(3pi)/4)#

#16cos(-(5pi)/12)#

Since #cos(-x)=cos(x)#, we have:

#16cos((5pi)/12)#

#16*(sqrt(2-sqrt(3)))/2#

#8sqrt(2-sqrt(3))#

#4.141#

As #4.141>0#, #4sin(4x-(3pi)/4)# is increasing at #x=pi/12#

Graphing #4sin(4x-(3pi)/4)#:

graph{4sin(4x-(3pi)/4) [-3.594, 3.334, -4.502, -1.04]}

We see that this is true.