What is the Domain of y=3e^(((x^2)-1)/(x))?
448269
The domain is all real numbers x, except for x=0.\n\nYou just have to remember that the exponential function and all polynomial functions are defined on the entire real line, that is there domain is all real numbers. Therefore, the only problem that occurs is with the operation of division by x. Since you cannot divide by 0, x can never be 0. All other values are fine though.\n\nIN RESPONSE to what urtimud said below: sure you can take e^0, but if you let x=0 you are taking e^(-1/0) which is undefined. Therefore, x=0 is not in the domain of this real valued function. Note that e is being raised to the ((x^2-1)/x) power, because your screen may have chopped off that last part.
766600
You have to use the quotient rule for this problem:\n\nFor any function f(x) defined by, f(x) = g(x)/h(x),\n\nf'(x) = [g'(x)h(x) - h'(x)g(x)]/[h^2(x)]\n\nIn this problem, we have:\n\ng(x) = xe^-x\nh(x) = x^3+x\n\nTherefore, \n\ng'(x) = xe^-x(-1) + e^-x(1) [you have to use product and chain rule for thie derivative]\n\nSimplfying, you get e^-x(1-x)\n\nh'(x) = 3x^2 + 1 (power rule)\n\nNow that we have all parts of the derivative, just plug in:\n\nf'(x) = [e^-x(1-x)(x^3+x) - (3x^2+1)(xe^-x)]/(x^3+x)^2\n\nNow just simplify this ugly mess =P\n\n---------\n\nHope this helps
894998
a)  x^2y^3\nb)  x^3/y^5\nc)  x^2/(y^4z^4)\nd)  x^2y^1/2 (assuming you mean y^-1/2 and not y-1/2)\ne)  x^2y^3t/[y^(4z6w)] (I'm assuming you mean y^(3t-4z-6w))
754001
the domain and the range of h(x):\nthe domain : x is less than or equal to 2,the range : y is greater than or equal to zero.\n\nDomain and range for the inverse of h(x) :\nD: x is grater than or equal to zero.\nR: y is less than or equal to 2.\nNotice that the range of h(x) is the same as the domain of   the inverse  of h(x).And the domain of h(x)is the same as the range of the inverse of h(x).
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).
1121129
Domain = all x-values included in the function\nRange = all y-values included in the function\n\nBoth will be written as an inequality.  Here's an example:\n\ny = square root(x-3) + 2\n\nGraph this equation on your calucluator or sketch a graph by hand by plugging in x coordinates starting with 3.  The vertex of this graph is at (3, 2).  Therefore, the domain is:\n\nx is greater than or equal to 3\n\nThe range is:\n\ny is greater than or equal to 2.\n\nIt really helps to visualize the problem when doing these problems.\n\nGood luck!
948239
xy = y^½\n y = y^½ /x\n y = y^½*x^-½\n\ndy/dx = -½ y^½ x^ -3/2\ndy/dx = - [½ y^½ ]/x^+3/2\ndy/dx = - (y^½) / (2x^3/2)
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
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)
979217
y = sqrt(1+t^2) = (1+t^2)^(1/2)\n\nby the power and chain rule\n\ndy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)\n\n= 2t(1/2) /sqrt(1+t^2)\n\n= t/sqrt(1+t^2)\n\nTherefore\n\ndy = (t / sqrt(1+t^2)) dt
931775
3e^x. Pull the constant outside the integral sign\nand just let it sit there!
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Domain:  All the values of x that provides a defined value of y.  In the equation above if we plug 0 or less than 0 for x, we get an undefined answer.  Thus the domain is x>0.  \n\nIntercepts:  To find intercepts we plug y equal to zero and then solve for x.  So, lnx/x = 0\n                          lnx = 0\n                             x = 1\nThus the intercept point is (x,y) = (1, 0)\n\nAsymptotes:  An asymptote is a straight or curved line which a curve will approach, but never touch.  By looking at the graph of lnx/x, we can conclude that the graph approaches y = 0 but never actually touches 0.  So, the asymptotes is y = 0.  \n\nCritical point:  In mathematics, a critical point (or critical number) is a point on the domain of a function where the derivative is equal to zero. It is also called a stationary point.  Derivative of above function is: \n                        dy/dx = [(x)(1/x)-lnx](x^2)\n                           0 = (1-lnx)/(x^2)\n                           0 = 1-lnx\n                           1 = lnx\n                            x = e\nSo, the critical point is x = e.
842257
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
1055222
Is that e^y?\n\nCompute the differential of both sides\n\n7dxe^y + 7xe^ydy + 9x^2dx = 20y^3dy\n(20y^3 - 7xe^y)dy = (7e^y + 9x^2)dx\ndy/dx = (7e^y + 9x^2)/(20y^3 - 7xe^y)
980558
∫xe^x dx \n= ∫x d(e^x)/dx dx \n= xe^x - ∫e^x (dx/dx) dx\n= xe^x - ∫e^x dx\n= xe^x - e^x + c\n= (x - 1)e^x + c
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|