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)).
