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)
= 0
(100t-160)/√(100t^2-320t+400) = 0
100t - 160 = 0
t = 160/100 = 8/5
etc
To take the derivative of the function, you can follow these steps:
Step 1: Identify the function
The given function is:
f(t) = sqrt(100t^2 - 320t + 400)
Step 2: Apply the power rule
To take the derivative of a function that contains a square root, you can apply the chain rule. However, before doing that, let's rewrite the function using exponential notation to make things easier:
f(t) = (100t^2 - 320t + 400)^(1/2)
Now, we can apply the power rule, which states that the derivative of x^n is n * x^(n-1).
So, for our function f(t), we have:
f'(t) = (1/2) * (100t^2 - 320t + 400)^(-1/2) * (200t - 320)
Step 3: Set the derivative equal to zero
Now that we have the derivative, we can set it equal to zero to find the critical points (where the slope is zero).
(1/2) * (100t^2 - 320t + 400)^(-1/2) * (200t - 320) = 0
Step 4: Solve for t
To solve for t, we can multiply both sides by 2 and divide both sides by (200t - 320). This will eliminate the fraction and give us:
(200t - 320) = 0
Step 5: Solve for t
Simplifying the equation, we have:
200t = 320
Dividing both sides by 200, we get:
t = 320/200
Step 6: Simplify
Simplifying, we have:
t = 8/5
So, to equate the derivative to zero, we solved for t and obtained t = 8/5.