what is the derivative of [(x^3)/2-x]^4/3; d/dx(f'(x))?
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
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.
183101
You can use the product rule the way it is written right now, or you can use foil and then take the derivative the regular way.  I think the second would be less work.\n(x+1)(x+3)\n= x^2 + 4x + 3\nThe derivative is 2x + 4\n\nIf you use the product rule, you would have\n(x+1)(x+3)' + (x+1)'(x+3)\n= (x+1)(1) + (1)(x+3)\n= x+1 + x+3\n= 2x + 4
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
1121843
the given function is not differentiable at x=0\n\ntherefore,   f(x) = x   for x>0        dy/dx = 1\n                           f(x) = -x  for x<0        dy/dx = -1
882357
The derivative of ln|f(x)| is 1/f(x) * d f(x) / dx;  this will be\n\n1/[sec5x + tan5x] * d(sec5x) / dx + d(tan5x) / dx\n\nd(sec5x) / dx = d(1/cos5x) / dx = -1/cos^2(5x) * -5sin5x = 5tanx*secx\n\nd(tan5x) / dx = 5sec^2(5x)\n\n(5tanxsecx + 5sec^2(5x)) / (sec5x + tan5x)\n\nFactor out 5sec5x to get 5sec5x * (tan5x + sec5x)/(sec5x + tan5x) = 5sec5x
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.
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
715822
Limit (X^3 -8)/(X^2 -4)\nX ->2\n-----------------------------\n\n(X^3 -8)/(X^2 -4) = (2^3 - 8)/ (2^2-4)= 0/0, which mean you can use L' Hospitals rule which is take the derivative of numerator and denominator separately\n\n3x^2/2x, and you plug in 2 and you get 12/4 = 3\n\n------------------------------------------------------------------\nLimit (3lnX)/(5(x-1))\nX ->1\n----------------------------------\n(3ln1)/(5(1-1)), that is the same as 0/0. (ln(1)) = 0 fyi\n\nso take the derivative just like the first one and you get \n\n(3/x)/5\n(3/1)/5= 3/5\n-------------------------------------------------------------------------\n\nLimit (e^x)/(X^3)\nX ->infinity\n--------------------------\n\n(e^(inf))/((infin)^3) = inf/inf\n\nThis one is undefined because as you keep on taking the derivative, the numerator is going to stay e^x and the bottom is going to be infinity until it is equal to 0 on the fourth derivative.\n\nSo the answer is undefined or does not exist
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.
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
805634
It is easier (by far) to do this with logarithmic differentiation:\ndy/dx=y*d/dx[ln y].\n\nHere, with y equal to your expression,\nln y=ln 10 +ln x +4ln(x-1)\n-3ln(x-2)-2ln(x+1),\nso the derivative of the logarithm is\n1/x +4/(x-1) -3/(x-2)-2/(x+1).\nThe first derivative is now your original expression times this.\n\nTo find the second derivative, use the product rule. The derivative of y is already known, and the second derivative of the logarithm is easy.
192120
Use the chain rule. Let f(y)=y^(1/2) and g(x)=x-2. Then your expression is just f(g(x)), and the chain rule states that D(f(g(x))) = Df(g(x))*Dg(x) = [(1/2)*(g(x))^(-1/2)] * [g'(x)] = [(1/2)*(x-2)^(-1/2)] * [1].\nI suggest doing several more of these questions until you no longer need to explicitly write out f and g, but can do so in your head.
757602
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
887032
This is a fundamental theorem of calculus problem.\n\nThe derivative of the integral of t * sqrt (t^3 + 2t) dt from -1 to x is\n\nx * sqrt (x^3 + 2x)
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
195265
lim h->0 (f(x+h)-f(x))/h\n\n[(x+h)^2-3(x+h)+1-(x^2-3x+1)]/h\n=(x^2+2xh+h^2-3x-3h+1\n    -x^2+3x-1)/h\n=(2xh+h^2-3h)/h\n=(2x+h-3)\n\nThere should be a limit as h->0 in front of each line above, so let h=0 in the final line to give you 2x-3.\n\nThis is called a derivative by the way, and if you haven't learned them yet there are easy rules to follow for computing the derivative of polynomials like this one.