What is the derivative of x^2?
347106
Okay, if you're actually serious, you REALLY should be paying more attention in class, but I'm gonna answer it for two points anyway.\n\nThe derivative of x^2 (with respect to x) is 2x.\n\nHave fun.
919994
Find the 2nd derivative and find the max.
440148
x=30\n\nRevenue is max when xp is max.\n\nxp=Rev=-0.2x^2 + 12x\n\nThe derivative is given by -0.4x+12 which = 0 when x = 30\n\nRev at x=30 is 180
750604
As Stuart has said above, you need to use the product rule to work this out.  The product rule states:\n\nthe derivative of f(x)g(x) is f(x)g'(x) + f'(x)g(x)\n\nSo, when you differentiate sinx wrtx you get cos and when you differentiate 2x wrtx you get 2.  Plugging these into the formula above you get:\n\n(f(x)g(x))'=2x(cosx)+2(sinx)\n                =2xcosx + 2sinx\n\nHope that this is helpful!
376765
Let u = 1 + x^4 and then apply the Chain Rule. This allows your expression to be written as\n\n1/sqrt(u)^(1/2)\n\nwhich is equivalent to\n\n1/u^(1/4)\n\nwhich, in turn, is equivalent to\n\nu^(-1/4).\n\nDifferentiating this expression with respect to x yields the following:\n\n(-1/4)*u^(-1/4-1)*(du/dx).\n\nNote: du/dx = 4x^3.\n\nTherefore, the desired derivative is \n\n(-1/4)*(1+x^4)^(-5/4)*(4x^3).\n\nIf we simplify this a bit we are given\n\n-x^3/(1+x^4)^(5/4).
814825
Integrating the derivitive leaves you with the initial equation x^2 since an integral is the opposite to a derivitive much like addition is the opposite of subtraction.  If you add 2 then subtract 2 you have the same answer.  As for the area under the curve, this will indeed be your entire population.
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
979434
What you've got there is a function of a function, so you need to use the old chain rule on it. (Remember also, that if F(x) = kf(x), where k is a constant, then F'(x) = kf'(x).)\n(I take it x^3 means "x cubed".)\nSo, given that the derivative of ln y is 1/y, this makes the derivateive of 2 ln x^3 = (2/x^3) *3 x^2, and you can tidy up a bit and get (6x^2)/x^3 which cancels down to 6/x.
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.
162293
On both equations, you have x^ but don't say what x was raised to. I'm assuming that it is x^2. Is this correct?
790977
The basic formula for differentiation of x^n is n.x^(n-1).\nSo your problem can be re-written as (x-1)^1/2.\n\nHere n is 1/2\n\nTherefore n.x^(n-1) gives (1/2).(x-1)^(1/2-1) (multiplied by 1)\n\nNow (1/2-1) is -(1/2).Therefore the entire expression becomes (1/2).(x-1)^(-1/2)\n\nNow this can be written as 1/{(2(x-1)^1/2)}\n\nThat is we are bringing that -1/2 down and it becomes positive.\n\nBut anything to the power of 1/2 is the root of that 'anything'.\n\nSo your final answer should be \n\n        1\n------------------\n2(sqrt of (x-1))\n\nRemember the differention of the bracket term,that is (x-1) is 1. As d/dx(x) is one and -1 is a constant. Hence that multiplied by one in the third sentence.
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
534670
Jimvalentinojr had the derivative right but didn't do enough algebra with it.  The x² he said was in the denominator was in the denominator of a fraction that was part of the denominator of another fraction, so the thing to do is simplify the complex fraction.  We had the derivative as\n\nΘ' = (1/x²)(-(b+a)/[1+(b+a)²/x²]) + (1/x²)(a/[1+a²/x²])\n\nso we multiply top and bottom of the 2 complex fractions by x² and get\n\nΘ' = (1/x²)[-x²(b+a)/(x²+(b+a)²)] + (1/x²)[ax²/(x²+a²)]\n\nand in each term x² cancels out, leaving\n\nΘ' = -(b+a)/[x²+(b+a)²] + a/(x²+a²)\n\nnow setting Θ' = 0,\n\na/(x²+a²) = (b+a)/[x²+(b+a)²], and cross multiplying (denominators never 0)\n\n(b+a)x² + (b+a)a² = ax² + a(b+a)²\n(b+a-a)x² = a(b+a)² - a²(b+a)\nbx² = ab² + 2a²b + a^3 - a²b - a^3\nbx² = ab² + a²b\nx² = ab + a²\nx = sqrt(ab + a²), and of course we ignore the negative root.
844269
assuming that your question is V(w)=w^2*(3/(8w^3)),\nwe can simplify it as V(w)=3w^2/8w^3 = 3/8w\nso, the derivative will be V'(w) = -1*3/8w^2 = -3/8w^2
586893
If its differentiate x * sqrt 1 + x^2 , the answer is\n  1 + 2x ( since sqrt 1 is a constant and d x^2/dx = 2x)\n\n \nIf your question is sqrt1 + x^2 answer is 0 + 2x = 2x
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)