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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?

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    $\begingroup$ $\frac{dy}{dt}=\frac{1}{\sqrt{2x+1}}\frac{dx}{dt}$ $\endgroup$ – Salech Rubenstein Mar 3 '12 at 22:46
  • $\begingroup$ I know that, I just typed it out wrong. $\endgroup$ – toby yeats Mar 3 '12 at 22:49
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    $\begingroup$ You're supposed to find $\frac{dy}{dt}$ I assume? Your derivative is wrong. I think you forgot to account for the derivative of $2x+1$ $\endgroup$ – Mike Mar 3 '12 at 22:53
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    $\begingroup$ Yes I get it now, the derivative of 2x+1 is 2*x prime $\endgroup$ – toby yeats Mar 3 '12 at 22:56
  • $\begingroup$ If you get it now, then write it up as an answer. If no one points out a mistake in your answer, then accept it. $\endgroup$ – Gerry Myerson Mar 4 '12 at 0:06
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$$ \frac {dy}{dt} = \frac {1}{2} (2x + 1)^{-1/2} 2* \frac{dx}{dt} $$

$$ \frac {dy}{dt} = \frac {1}{2} (9)^{-1/2} 2* \frac {dx}{dt} $$

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.

$$ \frac {dy}{dt} = \frac {1}{2} (9)^{-1/2} *6 $$

$$ \frac {dy}{dt} = \frac {1}{2} \frac {1}{3} *6 $$

$$ \frac {dy}{dt} = 1 $$

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