How do you calculate the third derivative of v=d/t?
84751
I presume you mean the third derivative of V(t,D) with respect to t, not D. The first derivative of V(t,D) with respect to t is acceleration a,\n \na=dV(t,D)/dt=-D/(t^2)\n\nThe second derivative of V(t,D) with respect to t is the boost B,\n\nB=da/dt=2D/(t^3)\n\nThe third derivative of V(t,D) with respect to t is\n\ndB/dt=-6D/(t^4)
761744
use the product rule, with the first term as t and the second as (4-t)^1/2\nproduct rule says that the derivative is second term times the derivative of the first term plus first term times the derivative of the second term.  Don't forget to use the chain rule along the way.\n\nThat is\nxy would have a derivate of yx' =xy'\n\nSo here\n[(4-t)^1/2](1) + (t){[1/2(4-t)^ -1/2][-1]\n(4-t)^1/2 + -t/2(4-t)^ -1/2
948100
t-(4/t)=3\ncan be rewritten as\nt^2 - 3t - 4 = 0\n\nthis has 2 possible solutions:\nt=-1\nor\nt=4
878312
V=4/3pier^3\nA=4pier^2
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
1043256
As you can tell from the various answers, your question is ambiguous.  If y is a function of x, then d(y^3)/dx = 3y²(dy/dx), using the chain rule, not + dy/dx as metaraison would have it.  If y is not a function of x, then dy/dx = 0 as if y were a constant.
853051
We have to find derivative of, g(x) = the integral (with lower limit tanx and upper limit x^2) of 1 / [sqrt(2+t^4)] dt.\n\nNow, the derivative of the integrand f(t)=1 / [sqrt(2+t^4)] is \nf'(t)= -2*t^3 / (2+t^4)^(3/2) [By Quotient Rule]\n\nd(g(x)) /dt= d {int(tan x)(x^2) {1 / [sqrt(2+t^4)] }dt} /dt\n=>dy /dt= d {f(t)}/dt (at tan x) - d {f(t)}/dt (at x^2)\n=>dy /dt= f'(t) (at tan x) - f(t) (at x^2)\n=>dy /dt= [2*t^3 / (2+t^4)^(3/2)] (at tan x) - [2*t^3 / (2+t^4)^(3/2)] (at x^2)\n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^2)^3 / (2+(x^2)^4)^(3/2)] \n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^8) / (2+(x^16))^(3/2)]\n\nI hope that answers your question.
1139627
dy/dx = 2x-3\nat X = 0. dy/dx = 2(0) -3 = -3
826981
It looks to me like it's .5d which is not too far from your answer.\n\nFor simplicity, use a unit of length in which d = 1 and a unit of time in which the speed of each of the two bugs is also 1. Let r be the distance between them and let h be the angle between the y direction and the vector from S to R. In other words h starts out pi/2 and approaches 0.\n\nThe time derivatives are:\n\ndh/dt = -sin(h)/r , dr/dt = -1 + cos(h)\n--->\ndr/dh = r(1-cos(h))/sin(h)\n\nSeparating the variables r and h to opposite sides of the equation and integrating both sides with the initial condition (r=1 at h=PI/2) gives:\n\nr = tan(h/2)/sin(h)\n\nYou want the distance between the two bugs as t-->oo or equivalently as h --> 0 which is r=1/2 (use L'hopital's rule for example), or d/2 in terms of the initial distance d.\n\nOf course S never actually reaches the y axis but approaches it asymptotically.\n\n(Maybe I should add, I first tried to formulate the problem in terms of the Cartesian coordinates and was able to write a differential equation for the curve but it looked very difficult to solve. Finally tried using the above coordinates (r,h) and it became much easier)
768228
a. V ice = pi(0.12^2-0.10^2)*3 = 0.041 m^3\nb. V ice = pi[(0.10 + x)^2 - 0.10^2]*3 = pi(0.10^2 + 0.20x + x^2 - 0.10^2)*3 = 3pi(0.20x + x^2)\ndV/dt = pi[0.60 + 6(0.02)]dx/dt = 0.000002pi\n= [0.60 + 6(0.02)]dx/dt = 0.000002\ndx/dt = 0.000002/[.60 + 6(.02)]
954124
a.dy/dt=(3/2)(0.2)t^1/2.\nwhen t=1\n=(3/2)(0.2)(1)^2/\n=0.3 cm/sec\nwhen t=2\n=(3/2)(0.2)(2)^1/2\n=0.4242 csps\n\nb.at t=3 it will be the thickest
764897
use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
660602
d/dx[y/x^2] gives y'/x^2 - 2y/x^3 on expansion
725593
dV/dr= 4/3pi*(3r^2)=4pi*r^2, evaluate this at r=3 you get 4pi*3^2 = 36pi.
184556
Let x(t) = A*cos( w*t + theta )\n\nthen x(0) = A*cos( theta )\nand dx/dt (0) = -A*w*sin(theta)\n\nso if you know x and its derivative in t=0 u have two equations and two unknowns ( A and theta ) . the rest is simple algebra .
274505
dy/du=6u^2\ndy/du=(2x^2)/3
794871
assuming 0 and 3 are limits of integration:\nv(t) = dx/dt = t^2 - 2t\nx = (1/3)t^3 - t^2 + C\nx = (1/3)3^3 - 3^2 - 0 + 0\nx = 9 - 9 = 0\n\nIf the limits are 0 and 2, \nx = 8/3 - 4 = -4/3\n\nGoing backwards,\n(1/3)t^3 - t^2 = 8/3\nt^3 - 3t^2 -8 = 0\nt ≈ 3.612887865