What's the domain of f(x) = x^x?
466400
0 can't be in the domain.\nAlso, -1/2 cannot be in the domain because\n(-1/2)^(-1/2) = 1/sqrt(-1/2) which is not a real number\nI'm not sure exactly how to right this:\nThe domain is any real number except 0 and any number of the form (p/q), where p and q do not have any common factors and q is even and (p/q) is negative.\nIf (p/q) is negative and reduced and q is even, then you would be taking an even root of a negative number(which is never a real number).
1090386
What?\nIf u=x^x\ndu/u=(1+ln(x))dx\n\nInteral  exp(x^x)dx=integral exp(u)du\n\nAnalytical solution for this indefinite integral is dificult.
634912
I think the problem you meant is:\n\n ((2X+1)^(1/3))/(X²-10X-11).  \n\nThe domain (for real numbers) is limited by two things:  Division by zero, and the even root of a negative number.\n\nHere the denominator can be factored into (X-11)(X+10) and that will equal zero at +11 and -10.  The cube root of a negative number is not a problem.  So the domain is all real numbers not equal to +11 or -10.
785253
So are you looking for a function with a singularity on (0,1)?  That would address the question I think.\nmaybe (x-(1/2))^(1/2)\n\nTHE FIRST ANSWER IS WRONG, I THINK, BECAUSE YOUR DOMAIN IS OPEN AND DOESN'T INCLUDE 0.
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
868237
x^(1/2).  that's x raised to the 1/2 power.
193948
The sq rt of x is x to the 1/2 power; 1 over sqrt x is x to the -1/2 power\n\nThen use your normal rule for exponents : d/dx (x^n) = n x^(n-1)\n\nd/dx x^-1/2 = -1/2 x^(-3/2)\n\n(This equals -1 over the square root of x cubed if you need to write it that way.)
1169084
Domain lists all possible x values. A normal polynomial function has a domain of all real numbers, because EVERY x value works.\n\nYour task is to find x values that DON'T work...and since you cannot divide by zero, then any x in the denominator that makes x^4+1 equal to zero will not be part of the domain.\n\nSo, what do you know about x^4+1?\n\nIt is an even power, so even if you put a negative number in for x, x^4+1 will be positive (since adding one will only make it more positive). If x=0, that is still positive, because of the plus one.\n\nx^4+1 is ALWAYS positive for any x...therefore it is NEVER equal to zero. Therefore any x value works, and your answer is all real numbers!!
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
824813
f(x) = x^m(1-x)^n\nm zeros at  x = 0 and n zeros at x = 1 \nSo x-axis is tangential to the curve at x = 0 and x = 1\n\nf'(x) = mx^(m - 1)(1 - x)^n + x^m * -n(1 - x)^(n - 1)\n= x^(m - 1) * (1-x)^(n - 1) * (m(1 - x) - nx)\n=  x^(m - 1) * (1-x)^(n - 1) * [m - (m + n)x]\n= 0 for stationary points.\n\nSo (m - 1) zeros at x = 0, (n - 1) zeros at x = 1 and a zero at x = m / (m + n)\n\nNow there has to be a turning point between x = 0 and x = 1 as f(x) is a polynomial and therefore continuous and differentiable for all x\n\nSo examine the possibilities\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... - .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... - .......... + ........... + ... Local minimum at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n ............... + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... + .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m .................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... + .......... + ........... + ... Local minimum at (1, 0)
750014
a. The domain is (0, ∞) This is found by noticing that since x is in the denominator, it can not equal zero, and also because ln(x) is only defined for x > 0.\n\nb. This is asking to solve the equation 0 = (lnx)/x. Since x can not equal zero, this is true when lnx = 0. This is easily found by solving the equation e^(lnx) = e^0. So whenever x = 1.\n\nc. I prefer to use the product rule instead of the quotient rule. So we can rewrite this equation as (lnx)(x)^(-1). So f' = gh'+hg', and g = lnx and h = x^(-1). So g' = (1/x) and h' = (-1/x^2). So f' = -(lnx)/(x^2) + (1/x^2) = (1 - lnx)/(x^2).\n\nd. This is increasing whenever f' > 0. So whenever [(1-lnx) > 0 and x^2 > 0] or [(1 - lnx)<0 and x^2 < 0]. Since x^2 is always > 0, because x > 0, we only need to worry about whenever (1 - lnx) > 0. Solving this inequality gives us whenever 1 > lnx. So, whenever x < e this inequality is true. That means that the function is increasing on the interval (0,e).\n\ne. Using our results from the previous question, and knowing that f is decreasing whenever f' < 0, we can see that this is true whenever 1 - lnx < 0 and x^2 > 0. Once again since x^2 > 0 for all x in the domain, we need only concern ourselves with 1 - lnx < 0. Solving this inequality gives us 1 < lnx. This is true whenever x > e. So the function is decreasing on the interval (e,∞).\n\nf. The relative maxima/minima are where f' = 0. Once again, we need only concern ourselves with when 1 - lnx = 0, since x^2 can not equal zero. So whenever 1 = lnx, or whenever x = e. Since we are not asked to classify this as a maxima or a minima, this answers the question.\n\nIf you want to know, it is a maximum and this can be determined using the second derivative.\n\ng. Now we will find the second derivatice, f''. Once again, I like to use the product rule. Since f' = (1-lnx)/x^2 we can say this is (1 - lnx)(x^(-2)). So g = (1 - lnx) and h = x^(-2) and g' = -(1/x) and h' = (-2/x^3). So f'' = (1 - lnx)(-2/x^3) + (-1/x^3) = (2lnx - 3)/(x^3).
387682
This expression is defined for every real number, domain is the whole set R, because it wouldn't be defined only for x^2+3=0, but x^2+3 is always greater than zero.
920193
e^(xlnx )=x^x\nderivative is  1+ln x
133829
8x^7\n\nThe first reference refers to this as the "Power Rule".\n\n  x^n'=nx^{n-1}
239640
Domain of a function f:X->Y is the set of all members of X. A well-defined function must map every element of the domain to an element of its codomain. In the given function, square root(x) is defined for all x if y can be a complex number. So,  if y belongs to the set of complex numbers, the domain is the set of complex numbers. And if, y is any real number, then x is the set of al positive real numbers. For more info, read http://en.wikipedia.org/wiki/Domain_%28mathematics%29 .
275631
One way to see this is to write x^(1-x) as exp((1-x)ln x) and use the chain rule. Only one person has the correct answer, art_colls. The power rule *does not apply* since the exponent is a function also!
952602
This is what always helps me.\n\nA function is a machine.\n\nThe domain is everything you can put into the machine so that it doesn't break.\n\nThe range is everything you can get out of the machine.\n\nIn the case of your function, there's only one possible value of x that will "break" the function, in the sense that you'd get an infinite answer.  That value is x=0, because then you'd have 9/0 + 7, and 9/0 is infinity.\n\nSo the range is {x | x is not equal to 0}.\n\nAssuming that you meant 9/x + 7, rather than 9/(x+7).  If you meant the latter, x cannot be equal to -7.