Second Derivative of [9√x – x^3/2] ?
931778
Since you have conflicting answers, I'm just chiming in to say go with ManiacalAce -- he got the correct answer.\n\nCorrect second derivative is as follows:\n\n(-9/4)x^(-3/2) - (3/4)x^(-1/2)\n\n~ ♥ ~
179824
Its preety easy. The answer is (3X^2)(3^X). Just take the derivative of X^3. Don't do any thing with 3^X cuz X is an exponent and when a X is an exponent you just leave it like that.
832910
yes the integral of the derivative of f(x)=x^2 is x^2.\n\nand just to check:\nf(x)=x^2\nf'(x)=2x\nF(x)=2*(1/2)x^(1+1)=x^2\n\nwhether you graph it or not makes no difference.
98350
limit as x tends to 3 ;( x^2 -3)/2 =(3^2 -3)/2 =(9-3)/2 = 3
754435
I | AUG is not correct.  You cannot differentiate power functions by the exponent rule.  (Even doing that, she made an error.)  The implicit differentiation done by Greg G is correct, or you can use the functional power rule:\n\nd(ƒ^g)/dx = (ƒ^g)*[(dƒ/dx)*(g/ƒ) + (dg/dx)*lnƒ]
739947
To do this you find the first derivative:\nf'(x) = 3x^2 -8x + 3\nWhen f'(x) = 0 there is a miinimum or maximum.\nWhere is this one increasing to a max and/or min?  Set it equal to 0,\n3x^2 -8x + 3 = 0; apply the quadratic formula to find the 0's for x:\n\nx = {-(-8) +- sqr[(-8)^2 - 4(3)(3)]}/2(3)\n   = {8 +- sqr[64-36]}/2 = {8 +- sqr[28]}/2 = {8 +- 2sqr(7)}/6\n   = (4+-sqr(7))/3 = x1, x2\n\nNow you have two values of x; now look at f(x) in the interval\n(-infinity, x1), then (x1,x2), then (x2,+infinity); this will tell you the answer.  Make a table showing the values approaching x1, \nx1 to x2, then x2 on.  And there you have it.
919994
Find the 2nd derivative and find the max.
745843
Use the quotient rule\nThe derivative is\n[(Q)(25000 +150 + 3Q)' - (25000 + 150 + 3Q)(Q)'] / (Q)^2\n= [(Q)(3) - (25000 + 150 + 3Q)(1)] / Q^2\n= [3Q - 25000 - 150 - 3Q] / Q^2\n= [-25000 - 150]/ Q^2\n= -25150/Q^2\nIf you want (and I would recommend this), then you can combine 25000 and 150 to get 25150.
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
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
739659
dy/dx=(3-√x)²/√x\n=(9+x-6*(x)^(1/2)/x^(1/2)\n=(9x^(-1/2)+x^(1/2)-6)\ndy=(9x^(-1/2)+x^(1/2)-6)dx\nintegrate\ny=int{9x^(-1/2)+x^(1/2)-6}dx\n  = 18x^(1/2)+2/3x^(1.5)-6x+D\nuse initial conditions\n2/3=18+2/3-6+D\n>>D= -12\n\ntherefore,\ny=2/3x^(1.5)-6x+18x^(1/2)-12\n\ni hope that this helps
87719
x^2 + 9x + 10x + 90\n\n(x + 9)(x + 10)
738083
3√x times 3√x, this means you multiply 3√x to 3√x\n3√x * 3√x = (3*3)*(√x * √x)\n                = (9)*((x^(1/2)) * (x^(1/2)))\n                = (9)*((x^[(1/2)+(1/2)])\n                = (9)*((x^[1])\n                = (9)*x\n                = 9x\nWhich is not 9√x.\n\nI hope this makes clear.
139329
calculus, fun stuff. To get the derivative, you need to drop the exponent then subtract one from the explonent. SO the X^2 would be 2x and 2x would just be 2 .    Final answer is 2x + 2.   Hm.. explained that poorly. In general, to find the derivative, Take the exponent on the X, drop it in front so its multipluing, than subract one from the exponent and have that number in the old exponents place (3x^3 would turn into 9x^2, 4x^2 would be just 8x) any number with X^1 is just the number, and just a number with no X is 0.
1022751
I'm working on the assumption that what you are saying is:\n\n(sqrt(x+3) - 2)/(x-1)\n\notherwise, it's easy.\n\nApply L'Hopital's rule. Take the derivative of the numerator and denominator then re-evalate the limits.\n\n\nd(sqrt(x+3) - 2)/dx = d((x+3)^(1/2) - 2) = 1/2(x+3)^(-1/2)\nd(x-1)/dx = 1\n\nlim x->1 1/2(x+3)^(-1/2) = 1/2(4)^1/2 = 1/4\n\nSo lim x->1 (sqrt(x+3) - 2)/(x-1) = 1/4
1076189
x√x=729\nx³=729²=531,441\nx=(531441)^(1/3)=81
350899
coshx\nand the derivative of coshx is sinhx