# derivative  9 results

derivative - rate at which the value y of the function changes with respect to the change of the variable x

A companys cost function is C(x) = 16x2 + 900 dollars, where x is the number of units. Find th
A companys cost function is C(x) = 16x^2 + 900 dollars, where x is the number of units. Find the marginal cost function. Marginal Cost is the derivative of the Cost function. [B]C'(x) = 32x[/B]

Derivatives
Free Derivatives Calculator - This lesson walks you through the derivative definition, rules, and examples including the power rule, derivative of a constant, chain rule

Determine whether the statement is true or false. If y = e^2, then y’ = 2e
Determine whether the statement is true or false. If y = e^2, then y’ = 2e e^2 is a constant, and the derivative of a constant is 0. So y' = 0 So this is [B]FALSE[/B]

Functions-Derivatives-Integrals
Free Functions-Derivatives-Integrals Calculator - Given a polynomial expression, this calculator evaluates the following items:
1) Functions ƒ(x).  Your expression will also be evaluated at a point, i.e., ƒ(1)
2) 1st Derivative ƒ'(x)  The derivative of your expression will also be evaluated at a point, i.e., ƒ'(1)
3) 2nd Derivative ƒ''(x)  The second derivative of your expression will be also evaluated at a point, i.e., ƒ''(1)
4)  Integrals ∫ƒ(x)  The integral of your expression will also be evaluated on an interval, i.e., [0,1]
5) Using Simpsons Rule, the calculator will estimate the value of ≈ ∫ƒ(x) over an interval, i.e., [0,1]

Hope it's okay to ask this here?
a) 1800 is the cost to run the business for a day. To clarify, when you plug in x = 0 for 0 candy bars sold, you are left with -1,800, which is the cost of doing business for one day. b) Maximum profit is found by taking the derivative of the profit equation and setting it equal to 0. P'(x) = -0.002x + 3 With P'(x) = 0, we get: -0.002x + 3 = 0 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=-0.002x%2B3%3D0&pl=Solve']equation solver[/URL], we get: x = 1,500 To get maximum profit, we plug in x = 1,500 to our [I]original profit equation[/I] P(1,500) = ? 0.001(1,500)^2 + 3(1,500) ? 1800 P(1,500) = -2,250 + 4,500 - 1,800 P(1,500) = \$[B]450[/B]

If 800 feet of fencing is available, find the maximum area that can be enclosed.
If 800 feet of fencing is available, find the maximum area that can be enclosed. Perimeter of a rectangle is: 2l + 2w = P However, we're given one side (length) is bordered by the river and the fence length is 800, so we have: So we have l + 2w = 800 Rearranging in terms of l, we have: l = 800 - 2w The Area of a rectangle is: A = lw Plug in the value for l in the perimeter into this: A = (800 - 2w)w A = 800w - 2w^2 Take the [URL='https://www.mathcelebrity.com/dfii.php?term1=800w+-+2w%5E2&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']first derivative[/URL]: A' = 800 - 4w Now set this equal to 0 for maximum points: 4w = 800 [URL='https://www.mathcelebrity.com/1unk.php?num=4w%3D800&pl=Solve']Typing this equation into the search engine[/URL], we get: w = 200 Now plug this into our perimeter equation: l = 800 - 2(200) l = 800 - 400 l = 400 The maximum area to be enclosed is; A = lw A = 400(200) A = [B]80,000 square feet[/B]

The function f(x) = e^x(x - 3) has a critical point at x =
The function f(x) = e^x(x - 3) has a critical point at x = The critical point is where the derivative equals 0. We multiply through for f(x) to get: f(x) = xe^x - 3e^x Using the product rule on the first term f'g + fg', we get: f'(x) = xe^x + e^x - 3e^x f'(x) = xe^x -2e^x f'(x) = e^x(x - 2) We want f'(x) = 0 e^x(x - 2) = 0 When [B]x = 2[/B], then f'(x) = 0

The function P(x) = -30x^2 + 360x + 785 models the profit, P(x), earned by a theatre owner on the ba
The function P(x) = -30x^2 + 360x + 785 models the profit, P(x), earned by a theatre owner on the basis of a ticket price, x. Both the profit and the ticket price are in dollars. What is the maximum profit, and how much should the tickets cost? Take the [URL='http://www.mathcelebrity.com/dfii.php?term1=-30x%5E2+%2B+360x+%2B+785&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']derivative of the profit function[/URL]: P'(x) = -60x + 360 We find the maximum when we set the profit derivative equal to 0 -60x + 360 = 0 Subtract 360 from both sides: -60x = -360 Divide each side by -60 [B]x = 6 <-- This is the ticket price to maximize profit[/B] Substitute x = 6 into the profit equation: P(6) = -30(6)^2 + 360(6) + 785 P(6) = -1080 + 2160 + 785 [B]P(6) = 1865[/B]

The height of an object t seconds after it is dropped from a height of 300 meters is s(t)=-4.9t^2 +3
The height of an object t seconds after it is dropped from a height of 300 meters is s(t)=-4.9t^2 +300. Find the average velocity of the object during the first 3 seconds? (b) Use the Mean value Theorem to verify that at some time during the first 3 seconds of the fall the instantaneous velocity equals the average velocity. Find that time. Average Velocity: [ f(3) - f(0) ] / ( 3 - 0 ) Calculate f(3): f(3) = -4.9(3^2) + 300 f(3) = -4.9(9) + 300 f(3) = -44.1 + 300 f(3) = 255.9 Calculate f(0): f(0) = -4.9(0^2) + 300 f(0) = 0 + 300 f(0) = 300 So we have average velocity: Average velocity = (255.9 - 300)/(3 - 0) Average velocity = -44.1/3 Average velocity = -[B]14.7 [/B] Velocity is the first derivative of position s(t)=-4.9t^2 +300 s'(t) = -9.8t So we set velocity equal to average velocity: -9.8t = -14.7 Divide each side by -9.8 to solve for t, we get [B]t = 1.5[/B]