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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?

  • 0
    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.2012-03-04

1 Answers 1

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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 $