This is a product of two functions. We have to use the product rule:
#y=sqrt(x-1)(x+1)=f(x)g(x)#
#f(x)=sqrt(x-1)=(x-1)^(1/2)=(x-1)^(1/2)#
#g(x)=(x+1)#
#f'(x)=1/2(x-1)^(1/2-1)=1/2(x-1)^(-1/2)(1)=1/(2(x-1)^(1/2))=1/(2sqrt(x-1))#
#g'(x)=1#
#dy/dx=f'(x)g(x)+g'(x)f(x)#
#dy/dx=1/(2sqrt(x-1))(x+1)+(1)(sqrt(x-1))#
#dy/dx=(x+1)/(2sqrt(x-1))+sqrt(x-1)#
#dy/dx=((x+1)+2sqrt(x-1)sqrt(x-1))/(2sqrt(x-1))#
#dy/dx=(x+1+2(x-1))/(2sqrt(x-1)#
#dy/dx=(x+1+2x-2)/(2sqrt(x-1))=(3x-1)/(2sqrt(x-1)#
Now we can multiply both top and bottom by #sqrt(x-1)# to normalize it:
#dy/dx=((3x-1)sqrt(x-1))/(2sqrt(x-1)sqrt(x-1))=((3x-1)sqrt(x-1))/(2(x-1))#
