Related rates and implicit differentiation I can get the proper answer, but I don't quite know why.
I am supposed to find $dy/dt$ for the function $y = \sqrt{2x +1}$ if $dx/dt = 3$ when $x=4$.
For the derivative I get $$ \frac {dy}{dt} = \frac {1}{2} (2x + 1)^{-1/2} \frac{dx}{dt},$$ which then gives me 
$$ \frac {dy}{dt} = \frac {1}{2} (9)^{-1/2} \cdot 3 \frac {dy}{dt} = \frac{1}{2}, $$
which is wrong. I can also do
$$ \frac {dy}{dt} = \frac {1}{2} (9)^{-1/2} \cdot 2 \frac {dx}{dt},$$
which gives me $1$, which is the proper answer, but I am not sure why I get that. I know that the derivative of the inner function will be $2$ but the problems defines it as being $3$, so do I just multiply the two?

$$ \frac {dy}{dt} = \frac {1}{2} (2x + 1)^{-1/2}  2* \frac{dx}{dt} $$
$$ \frac {dy}{dt} = \frac {1}{2} (9)^{-1/2} 2* \frac {dx}{dt} $$
The 2 comes from the derivative of the inner function and then I multiply that by the implicit derivative of x which was given as 3 so I get 6.
$$ \frac {dy}{dt} = \frac {1}{2} (9)^{-1/2} *6  $$
$$ \frac {dy}{dt} = \frac {1}{2} \frac {1}{3} *6  $$
$$ \frac {dy}{dt} = 1 $$
