What are the first and second derivatives of a Surge Function? (product rule)?
Use product rule d(uv)/dx=udv/dx+vdu/dx u=8t v=exp(-0.45t) du/dx=8 dv/dx=-0.45exp(-0.45t) y'=-3.6texp(-0.45t)+8exp(-0.45t) Use the same method for the second derivative y''=1.62texp(-0.45t)-7.2exp(-0.45t)
How do I find the derivatives of these 3 functions?
1) f(x)=( cos²x)/2 f '(x)= 1/2 * d/dx ( cos²x) = 1/2 * 2cos x * d/dx( cos x) = cos x* (- sin x) = - sin x cos x OR = -(sin 2x)/2 2) f(x) = sin²x f'(x)= d/dx(sin²x) = 2sin x * d/dx( sinx) = 2 sin x * cos x = sin 2x 3) f(x)= -1/4 * cos 2x f'(x)= -1/4* d/dx ( cos 2x) = -1/4 * (- sin 2x) * d/dx( 2x) = 1/4 * sin 2x * 2 = 1/2 * sin 2x = ( sin 2x )/2 (Ans)
Find the second derivative of the function y= square root of (2x-1)?
To find the first derivative, you need the chain rule y = (2x - 1)^(1/2) y' = (1/2)(2x - 1)^(-1/2) * 2 y' = 1/[2sqrt(2x - 1)] * 2 y' = 1/sqrt(2x-1) (the 2's cancel each other. To find the 2nd derivative, you need the chain rule again y' = (2x - 1) ^ (-1/2) y" = (-1/2)(2x - 1)^(-3/2) * 2 y" = -1/(2x - 1)^(3/2) (the 2's cancel again)
How do you check if the second derivative of a function is positive, zero or negative without getting the exact value
Find its zeros and discontinuities, and determine if it is positive or negative between those intervals, exactly like you would any function.Alternatively, if you have a graph of the derivative, you can tell by looking at its slope, just like you can tell if a derivitive is positive or negative by looking at the original graph.Alternatively, on the original functions graph, see where the slope is concave up or down.It all depends on which information you have available; sometimes you are looking at the second derivative to properly graph the behavior of the function and sometimes you are looking at the function to determine the properties of a second derivative.
If the second derivative of a function equals zero at a specific point, is the point necessarily an inflection point?
Typically, yes, but not always. Typically, the second derivative is positive on one side of that point and negative on the other, and in that case, there’s an inflection point. But the second derivative could have the same signs on both sides of that point, and it wouldn’t be an inflection point.The simplest example is [math]f(x)=x^4[/math] whose second derivative is [math]f’’(x)=12x^2[/math]. The second derivative is [math]0[/math] at [math]x=0[/math], but positive on both sides of [math]x=0[/math]. There’s no inflection point there.
Let function f(x), 1st derivative, and second derivative all the positive on closed interval [a.b]?
Let the function f(x), its derivative, and its second derivative all the positive on a closed interval [a.b]. The interval [a,b] is partitioned into n equal length sub-intevals and these are used to compute an Upper Sum, U, a Lower Sum, L, and the Trapezoidal Rule a pprosimation, T. If I=Integralf(x)dx with limits a to b which statement is true a) L
Does anyone know the derivatives of these functions?
Product Rule: h(x) = f(x)g(x) h'(x) = g(x)f '(x) + f(x)g'(x) Chain Rule: h(x) = f(g(x)) h'(x) = g'(x)f '(g(x)) h(x) = 5xe^(6x) h'(x) = 5e^(6x) + 30xe^(6x) h'(x) = 5e^(6x)(1 + 6x) h(x) = x^(2)e^(3x) h'(x) = 2xe^(3x) + 3x^(2)e^(3x) h'(x) = xe^(3x)(2 + 3x)
What is the derivative of log 2x, log 3x, and log 4x?
Note these all differ by just a constant, [math]\log(nx) = \log n + \log x[/math]. So all of these have the same derivative equal to [math]1/x[/math].
How would you find the second derivative and velocity of these functions?
1) Find F''(2) if F(x) = x^2f(5x) and it is known that f(10) = −2, f '(10) = 1, and f ''(10) = −1. F''(2) = ______? 2) A body moves along a coordinate line in such a way that its position function at any time t is given by s(t) = t sqrt6 − t^2 0 ≤ t ≤ sqrt6 where s(t) is measured in feet and t in seconds. Find the velocity and acceleration of the body when t = sqrt3/2 Here's an image of the problem if needed: imageshack.us/photo/my-images/14/1618669... velocity = ____ ft/s acceleration = ____ ft/s^2 3) In a test flight of McCord Aviation's experimental VTOL (vertical takeoff and landing) aircraft, the altitude of the aircraft operating in the vertical takeoff mode was given by the position function h(t) = 1/128t^4 − 1/2t^3 + 8t^2 0 ≤ t ≤ 32 where h(t) is measured in feet and t is measured in seconds. Here is an image of the problem if needed: http://imageshack.us/photo/my-images/547/42040991.png/ (a) Find the velocity function. v(t) = ______ (b) What was the velocity of the VTOL at t = 0, t = 16, and t = 32? At t = 0 _____ ft/s At t = 16 _____ ft/s At t = 32 _____ ft/s (c) What was the maximum altitude attained by the VTOL during the test flight? ________ ft