We are given #y=(x-2)^2(x+1)(x+4)#
To find the gradient we must differentiate, but as our equation is in the form #y=u*v*w#, where #u#, #v# and #w# are functions of #x#. We must use the product rule where:
#(dy)/(dx)=(u'*v*w)+(u*v'*w)+(u*v*w')#
In this case, #y=u^2*v*w#, where:
#u=(x-2)#,
#v=(x+1)#, and
#w=(x+4)#
#(dy)/(du)=2u#
#(dy)/(dv)=1#
#(dy)/(dw)=1#
#(du)/(dx)= 1#
#(dv)/(dx)= 1#
#(dw)/(dx)= 1#
So, #(dy)/(dx)=2((x-2)(x+1)(x+4))+(x-2)^2(1)(x+4)+(x-2)^2(x+1)(1)#
Putting our values for #x=1# in gives us #2(1-2)(1+1)(1+4)+(1-2)^2(1+4)+(1-2)^2(1+1)=2(-1)(2)(5)+(-1)^2(5)+(-1)^2(2)=-13#
As #(dy)/(dx)# is negative when #x=1#, the gradient is negative and therefore decreasing.
