.
#y=cos^3(2x)+sin(3x)#
We have to take the derivetive of each term separately:
Let's do #cos^3(2x)# first:
Let's say #y_1=cos^3(2x)#. Note that this is #cos# of another function which is #2x#.
Let's put #u=2x#. Then #du=2dx#
#(du)/dx=2#
#cosu# itself is another function.
Let's say #z=cosu#. Now, let's take the derivative:
#dz/(du)=-sinu#
Also, plugging #cosu# into the original function give us:
#y_1=z^3#
#dy_1/dz=3z^2#
The Chain Rule says:
#dy_1/dx=dy_1/dz*dz/(du)*(du)/dx# Let's plug them in:
#dy_1/dx=(3z^2)(-sinu)(2)# Now, let's substitute back for #z#:
#dy_1/dx=(3cos^2u)(-sinu)(2)# Now, let's substitute back for #u#:
#dy_1/dx=3cos^2(2x)(-sin(2x))(2)# Now, we simplify:
#dy_1/dx=-6cos^2(2x)sin(2x)#
Now, we do the second term:
Let's say #y_2=sin(3x)# But #3x# is another function.
Let's say #w=3x# Then #(dw)/dx=3# and
#y_2=sinw#
#dy_2/(dw)=cosw#
The Chain Rule says:
#dy_2/dx=dy_2/(dw)*(dw)/dx# Now, let's plug them in:
#dy_2/dx=(cosw)(3)=3cosw# Now, let's substitute back for #w#:
#dy_2/dx=3cos(3x)#
According to our own definition:
#y=y_1+y_2# Then:
#dy/dx=dy_1/dx+dy_2/dx#
Now. we can plug the derivatives of both terms into the above equation:
#dy/dx=-6cos^2(2x)sin(2x)+3cos(3x)#
