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April 19, 2014

April 19, 2014

Posted by **Matt** on Tuesday, March 2, 2010 at 9:38pm.

sqrt of (100t^2-320t+400).

- Calculus AB -
**Heather**, Tuesday, March 2, 2010 at 10:52pmto take the derivative of an equation involving the square root it would be easier to set the whole equation in parenthesis and raise it to (1/2). you then bring the exponent down in front and decrease the by 1 to get (-1/2). this will make your equation in the denominator and times by (1/2). you then will have to take the derivative of the inside piece.

1/(2(sqrt(100t^2-320t+400))*(200t-320) which is the derivative of the inside piece)

and i'm guessing that by "equate" you mean set equal? sorry, i haven't heard many people say it that way. to do this just set the equation to zero and solve.

- Calculus -
**Reiny**, Tuesday, March 2, 2010 at 11:51pmHeather's derivative is not correct ...

y = (100t^2-320t+400)^1/2

dy/dx = (1/2)(100t^2-320t+400)^(-1/2)(200t - 320)

= 0

(100t-160)/√(100t^2-320t+400) = 0

100t - 160 = 0

t = 160/100 = 8/5

etc

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