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.
How do you take the derivation of the following function and equate to zero?
sqrt of (100t^2-320t+400).
2 answers
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)
= 0
(100t-160)/√(100t^2-320t+400) = 0
100t - 160 = 0
t = 160/100 = 8/5
etc
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
3 answers