So we know that

#sqrt(x+2)*sqrt(x+2) = x + 2#

Let's say that the first root is #f(x)# and the second is #g(x)#, so we have

#f(x)*g(x) = x +2#

Derivating both sides we have

#lim_(h rarr 0)(f(x+h)g(x+h) -f(x)g(x))/h = 1#

Now, since #f(x+h)g(x) - f(x+h)g(x) = 0# we can put that in a sum without changing anything, so

#lim_(h rarr 0)(f(x+h)g(x+h) -f(x+h)g(x) +f(x+h)g(x) -f(x)g(x))/h = 1#

Put #g(x)# and #f(x+h)# in evidence

#lim_(h rarr 0)f(x+h)*(g(x+h) -g(x))/h +lim_(h rarr 0)g(x)*(f(x+h) -f(x))/h = 1#

Evaluate the limit of the factors we just put in evidence

#f(x)lim_(h rarr 0)(g(x+h) -g(x))/h +g(x)lim_(h rarr 0)(f(x+h) -f(x))/h = 1#

The remaining limits are the definitions of #f^'(x)# and #g^'(x)#, so we can rewrite it to be

#f(x)g^'(x) + g(x)f^'(x) = 1#

(This is actually called the product rule and is widely used for more complex functions)

But since #f(x) = g(x)# and #f^'(x) = g^'(x)# so we can further rewrite to be

#2f(x)f^'(x) = 1#

Isolate #f^'(x)# and since #f(x) = sqrt(x+2)# put that back in.

#f^'(x) = 1/(2sqrt(x+2))#