need help on derivatives??
958491
i think we can convert the log with subsrcipt 3 to natural log\nlogx (with subscript 3) = lnx/ln3\nso derivative is\n1/(xln3)\nhope this helps
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.
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
773933
1-In general the derivative of ln|U| is U'/U. In your case U = 2x^x + 4x + 1. Using the fact that x^x = exp(xlnx) you can easily check that the derivative of \n\nU' = 2(1 + lnx)x^x + 4.\n\nThus the derivative of your function is\n\n[2(1+lnx)x^x+4]/[2x^x+4x+1].\n\n2- The derivative of U^a, where a is a real number, is aU^{a-1}U'. In your case U = x e^x and a=1/2. Since U' =(x+1)e^x, one has\n\n(sqrt(xe^x))' =1/2{(x+1)/sqrt(xe^x)}e^x = 1/2{(x+1)/sqrt(x)}exp(x/2).
1134715
Valuation of financial derivatives (options, futures, forwards, etc)\nReal hard math, but real good money.
1111738
the best way to appr this problem is to apply the first principle i.e.\nf'(x)= Lt     [f(x+h)-f(x)]/h\n       h->0\nnow putting f(x)=e^x we have\nf'(x)= Lt    [e^(x+h)-e^x]/h\n       h->0\n     =Lt    [e^x(e^h-1)]/h\n       h->0\n\nnow (e^h-1)/h where h tends to 0 is 1\nhence f'(x) = e^x\ntherefore the derivative of e^x is e^x
750296
-2x + h\nIn about 5 minutes you will be taken the limit of the expression as h ---> 0.  Then, you will be told that then it will be the derivative.\n\nAfter that you take 2 - x^2 do the derivative and get -2x.
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
786389
The derivative is [(x-4)- (x+1)]/(x-4)^2 = -5/(x-4)^2\n\nIf we use the definition of the derivative:\n\n\ndy/dx = Lim h -> 0 {f(x+h) -f(x)}/h\n= lim h-> 0 [(x+h+1)/(x+h-4) - (x+1)/(x-4)]/h\n                   (x+h+1)(x-4)-(x+1)(x+h-4)\n= lim x-> 0 ---------------------------------------\n                   (x+h-4)(x-4)h\n\n= lim h-> 0 -5h/(x+h-4)(x-4)h = -5/(x-4)^2
1060074
d (ln(x)) / dx = 1/x\n\nThe derivative of the natural log of x with respect to x equals 1 / x
99672
honestly....good luck\n\ntry www.mathworld.com\n\ndo a search for differential equations, i think that would be your best bet...but that can get confusing...good luck
921060
In general if you have x^n, the derivative is n * x^(n-1).\n\nFor x² the answer is 2x^1 or just 2x.\n\ny = x²\ndy/dx = 2x
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
316594
The anti-derivative of x^(2)*e^(x)^(3) is (1/3)*e^(x)^(3)\n\nYou can always check by taking the derivative of your answer.
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
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]
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.