Derivative of a Product

$F(x)=(x+2)*\sqrt { x-2 } \\ F(x)=(x+2)*{ (x-2) }^{ \frac { 1 }{ 2 } }\\ \\ Now\quad we\quad take\quad the\quad derivative\\ f(x)=(x+2)\quad and\quad g(x)={ (x-2) }^{ \frac { 1 }{ 2 } }\\ \\ F'(x)=f'(x)*g(x)+f(x)*g'(x)\\ F'(x)=1*{ (x-2) }^{ \frac { 1 }{ 2 } }+(x+2)*(\frac { 1 }{ 2 } { (x-2) }^{ -\frac { 1 }{ 2 } })*\quad 1\\ F'(x)=\sqrt { x-2 } +\frac { x+2 }{ 2\sqrt { x-2 } } \\ Now\quad we\quad substitite\\ F'(11)=\sqrt { 11-2 } +\frac { 11+2 }{ 2\sqrt { 11-2 } } \\ F'(11)\quad =\quad \sqrt { 9 } +\quad \frac { 13 }{ 2\sqrt { 9 } } \\ F'(11)\quad =\quad 3+\frac { 13 }{ 6 } \\ F'(11)\quad =\quad \frac { 18+13 }{ 6 } \\ F'(11)\quad =\quad \frac { 36 }{ 6 }$

f(x)= (x+2)*(x-2)^.5 --------eq. 1

Let (x-2)^.5=y That makes x-2=y^2 which makes x+2=y^2+4

Taking the derivative of this relation between gives us dx=2y*dy Hence, dy/dx= 1/2y

The derivative of y is 0.5/(x-2)^.5

Substituting y in equation 1 gives you f(x)= (y*(y^2+4) = y^3+4y

Now finding the derivative of f(x) gives us f'(x) =(3y^2 + 4) (dy/dx) Substituting x back gives f'(x)=(3(x-2)+4) (1/(2(x-2)^.5)) Putting x=11 we get 31/6

If y = uv then apply UV rule --> u * dv/dx + v * du/dx

f'(x)=3x-2/(2*((x-2)^1/2)