What is the antiderivative of f'(x)=cos(x^2)?
821024
∫cos(x^2) dx = √(π/2)FresnelC(√(2/π)x)
880391
f'(x)= cosx -sinx =0\nwhen cosx=sinx, \nand this happens when x= pi/4,\nnow\nf(pi/4) =  sin(pi/4) + cos(pi/4)=1/sqrt(2) + 1/sqrt(2) = sqrt(2) =1.414\nf(0)=sin(0)+cos(0) =1\nf(pi/3)= sin(pi/3) + cos(pi/3) =0.866 + 0.5 =1.366\nthe maximum is sqrt(2) and it occurs when x= pi/4\nthe minimum is 1 and it occurs when x= 0\n`
255922
Im not sure if you need clarification on how to do this. Ill provide a few details on how to get these answers:\n\n1) ∫sinxdx = - cos(x) + C\n-This one is relatively straightforward, just think "what gives me sin(x) when I take its derivative"\n\n2)∫e3xdx = (Im assuming this means e^(3x))\n         = (e^(3x))/3 + C\n-Just think of the chain rule backwards. If you take the derivative of e^3x you'll have to multiply by 3, so when you take the derivative divide by 3.\n\n3)∫cos3xdx = (Im assuming this is (cos^3)x)\n           = ∫((cos^2)x )(cos(x))\n           = ∫(1-(sin^2)x)((cos(x))\n      Now let u=sin(x)\n             du=cos(x)dx\n           = ∫(1-u^2)du\n           = [u - (u^3)/3]\n     Sub back in for u\n           = sin(x) - (sin^3(x))/3 + C\n-Wow, that one was hard to type out. I hope you can follow it somewhat. The trick is to write cos^3(x) as cos^2(x) and cos(x). Then the identity cos^2(x) + sin^2(x)=1 is used. The u substution allows us to find the integral at the end.\n\n4) Derivative of 3x^4 + 2x^2 + x + 3\n                  = 12x^3 + 4x + 1\n-The derivative is defined by some long equation, but you can use the following fact:\nIf f(x)=x^n\nthen f'(x)=nx^n-1\n\nHopefully you understand most of what I did. Im sure someone else will probably explain a lot of this in much better detail. Like I said, I wasnt sure how much detail you were looking for. \n\nPS Ill tell you site that will help you forever in any integral calculus course: http://integrals.wolfram.com/index.jsp
828677
∫sin(ln(t)) dt\n\n= ∫sin(ln(t)) d(t)/dt dt\n\n= tsin(ln(t)) - ∫t * 1/t cos(ln(t)) dt \n\n= tsin(ln(t)) - ∫ cos(ln(t)) dt \n\n= tsin(ln(t)) - ∫ cos(ln(t)) d(t)/dt dt \n\n= tsin(ln(t)) - (tcos(ln(t)) -∫ t * 1/t *-sin(ln(t)) dt)\n\n= tsin(ln(t)) - tcos(ln(t)) -∫sin(ln(t)) dt\n\nSo 2∫sin(ln(t)) dt = tsin(ln(t)) - tcos(ln(t))\n\nSo ∫sin(ln(t)) dt = t/2(sin(ln(t)) - cos(ln(t)))
861008
NO\n\nf'(x)=(-sinx)(e^cosx)\nf'(4.0)=(-.07)(2.7)= -1.89
829160
y'=2 cos(7x)*(-sin 7x)*7 By the Chain Rule
945648
Take the derivative of f :\nf'(x) = 5x^4 (x+7)^4 + 4 (x+7)^3 x^5\nf'(x)= x^4 (x+7)^3 [ 5 (x+7) + 4x]\nf'(x)= x^4 (x+7)^3 [ 9x + 35]\n\nSo the derivative is equal to 0 for the values :\nx= -7\nx= - 35/ 9\nx = 0\n\nthe function is increasing in [-11;-7],\nthe function is increasing in [-35/9;+12],
506639
The first derivitive of f(x) = 5/x^3 - 3/x^1/3 is...\n\nf'(x) = -15/x^4 + 1/x^(4/3)\n\nHope this helps. I used the product rule. Remember, you can write the term 5/x^3 and 5*x^-3. I find it simpler to rearrange the equation and use the product rule to get at the first derivitive.
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
920193
e^(xlnx )=x^x\nderivative is  1+ln x
986880
Do you want how to derive them or what the derivatives are?\n\nF(x)=a^x  ---> f'(x) = a^x (ln a)\n\nf(x) = sin x ---> f'(x) = cos x\n         cos x      --->           -sin x\n         tan x       ---->           sec^2 (x)\n         sec x      ---->           sec x tan x\n         csc x      ---->           - csc x cot x\n         cot x       ---->            - csc^2 (x)
304641
Integral f(x)dx=-3cos(pi)+3cos(0)=3\nSo average value of the func f(x) = 3x sin (x^2) is \n3/sqrt(pi)
209885
well if its dealing with derivitives\n\nf(x) = x^2\nf'(x) = 2x^1\n\nor the general rule\nf(x) = x^n\nf'(x) = n(x^(n-1))
872722
F'(x) = √(5-x) - x/[2√(5-x)]\n\nF''(x) = -1/[√(5-x)] - 1/[2√(5-x)] + x/[4√(5-x)]³\n= - 3/[2√(5-x)] + x/[4√(5-x)³]
870571
F(x)=2[sec(x)^2] * [tan(x)] / [xcos(x)+2sin(x)]\n\nremember:\nsec(x)=1/cos(x); tan(x)=[sin(x)/cos(x)]\n\nWe rewrite the function and get:\nF(x) = [2/cos(x)^2] * [sin(x)/cos(x)] / [xcos(x)+2sin(x)] <=>\n\nF(x) = [2 /cos(x)^2] * [sin(x) /{ cos(x) * [xcos(x)+2sin(x)]} <=>\n\nF(x)= 2sin(x) / [ x*cos(x)^4 + 2sin(x)cos(x) ] =>\n\nLimF(x)= 2sin(x)/[x+2sin(x)] \nx→0\nBecause all terms with cos(x)→1 IF x→0 =>\n\nLimF(x)=2sin(x)/[x+2sin(x)] = 2sin(x)/x \nx→0\nbecause 2sin(x)→0 IF x→0 =>\n\nLim F(x)=2 because sin(x)/x →1 IF x→0 \nx→0\n=============================\nRegard from Denmark
899944
first, I would put it in an easier to express format g(t)=[t^(1/2)](1-t)  then take the derivative g'(t)=[(1/2)t^(-1/2)](1-t)-t^(1/2)   now just solve for g'(t)=0, so 0=[(1/2)t^(-1/2)](1-t)-t^(1/2)      now we distribute to get 0=[(1/2)t^(-1/2)]-(3/2)t^(1/2)     we add (3/2)t^(1/2) to both sides to get (3/2)t^(1/2)=(1/2)t^(-1/2)   simplfy this to [3sqrt(t)/2]=[1/2sqrt(t)]   we multiply both sides by 2sqrt(t) to get 6t/2=1 or 3t=1, which means that t=1/3\n\nAnd by the way, this isn't the chain rule, this is the Product Rule
71756
Here's a classic:\n\nUsing the following partial deriviatives:\n\nInt(f(x)*g(x))dx = f(x)*G(x) - Int(f'(x)*G(x))dx\n\nwhere G(x): the primitive function of g(x)\n     and f'(x): the derivative of f(x)\n\nNow watch the following 'proof':\n\n    Int(1/x^2 * 2x)dx  =  1/x^2 * x^2 - Int(-2/x^3 * x^2)dx\n           {just take f(x) = 1/x^2 and g(x) = 2x}\n\n    this yields:\n\n    Int(2/x)dx  =  1 - Int(-2/x)dx  = 1 + Int(2/x)dx    \n\n    substracting Int(2/x)dx on both sides yields:\n\n    0 = 1