posted by Matt .
How do you take the derivation of the following function and equate to zero?
sqrt of (100t^2-320t+400).
to 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.
Heather's derivative is not correct ...
y = (100t^2-320t+400)^1/2
dy/dx = (1/2)(100t^2-320t+400)^(-1/2)(200t - 320)
(100t-160)/√(100t^2-320t+400) = 0
100t - 160 = 0
t = 160/100 = 8/5