Question

Square root of (x+deltaX)+ 2 - square root of (x+2)/ delta x

Answer 1

\[\frac{ \sqrt{(x+\Delta)+2} - \sqrt{x+2}}{dx}\]
multiply the top and bottom by the conjugate (change the minus sign to a plus)
then take the limit as Delta goes to zero.

Answer 2

\[\sqrt{[x+Deltax]+2}-\sqrt{x+2}*\sqrt{[x+Deltax]+2}+\sqrt{x+2}?

Answer 3

Yes. it's an ugly version of (a-b)*(a+b)= a^2 - b^2

Answer 4

is there simpler way?

Answer 5

would it be [x+delta+2]-[x+2]

Answer 6

the delta x cancels out and ur left with 1/sqrt[[x+deltax]+2]-sqrt[x+2]?

Answer 7

except when I cut and paste, I did not change the minus sign to a plus sign:
1/sqrt[[x+deltax]+2] + sqrt[x+2]

Answer 8

when delta x goes to zero you end up with
\[\frac{1}{2\sqrt{x+2}}\]

Answer 9

Fixed
\[ \frac{ x+\Delta x+2 - (x+2)}{\Delta x\sqrt{(x+\Delta x)+2} + \sqrt{x+2}} \]

Answer 10

got it thanks i forgot to plug in zero for delta x ur a life saver