The answer is: #y'=((5x-2)(x-2))/(2sqrt(x-3))#.

The rule of the derivative of a product is:

#y=f(x)*g(x)rArry'=f'(x)*g(x)+f(x)*g'(x)#;

so:

#y'=2xsqrt(x-3)+(x^2+4)*1/(2sqrt(x-3))=#

#=(2xsqrt(x-3)*2sqrt(x-3)+x^2+4)/(2sqrt(x-3))=(4x(x-3)+x^2+4)/(2sqrt(x-3))=#

#=(5x^2-12x+4)/(2sqrt(x-3))#

We can factor #5x^2-12x+4# remembering that if we consider that polynomial like a square equation, we can find the two solutions #x_1andx_2#, than:

#ax^2+bx+c=a(x-x_1)(x-x_2)#;

so, we can use the reduced formula because #b# is even:

#5x^2-12x+4=0rArr# #Delta/4=(b/2)^2-ac=36-20=16# and

#x_(1,2)=(-b/2+-sqrt(Delta/4))/a=(6+-4)/5rArr#

#x_1=2/5andx_2=2#.

So:

#5x^2-12x+4=5(x-2/5)(x-2)=(5x-2)(x-2)#.

Finally:

#y'=((5x-2)(x-2))/(2sqrt(x-3))#.