Is B^(-1/3) * C^(1/3) equivolent to (C/B)^(1/3)?
91593
yes.as b has something raised to power negative.it goes to denominator
726082
Yes since ∫f(x) dx = ∫x dx\n= ½x² + c where c is ANY constant.
43939
is the above question in first degreee derivative??
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
936251
I think f^(-1) (x) means the inverse of f(x), not the reciprocal.\n\nLey y =f(x)\ny = (x^3 +1)^(1/3)\n\nsubstitute the x and y\nx = (y^3 +1)^(1/3)\n\nnow solve for y\nx^3 = (y^3 +1)\ny^3 =x^3 -1\ny =(x^3 -1)^(1/3)\n\nNow if you put 1.5 into this equation\ny = ((1.5)^3 -1) ^(1/3)\ny =(2.375)^(1/3) or approx. 1.3342
388069
This is with a calculator or Excel, right?\n\nYour calculator probably attempts to calculate x^(1/3) by taking the natural log of x:\n\nx^(1/3) = exp(ln(x)*(1/3))\n\nbased on the fact that:\n\nln(a^b) = b * ln(a)\n\nbut you can't take the log of a negative number.  It doesn't recognize the fact that the cube root is well-defined.\n\nYou can help it along by substituting:\n\nsign(x)*(abs(x))^(1/3) = x ^ (1/3)\n\nbasically taking the sign out at the beginning and putting it back at the end.
783756
in standard form yes, but thats not the whole answer, the answer would be x to the 2 power and -1
546126
intrgrate o e^(3x)=(1/3)e^(3x)
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
907483
1/x^2 = x^(-2)\nThe antiderivative is x^(-2+1)/(-2+1) + C\n= x^(-1)/-1 + C\n= -1/x + C\nIt is similar for 1/x^3\nYou do partial integrals the same way you do regular integrals.  You need to know which one you are integrating with respect to and then pretend the other variable(s) are constants.
697108
e^(1+lnx) = e^1 * e^lnx = e*x = ex\nder(ex) = e+c
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
961234
Let x = y^(1/3)\nx^2 - 9x + 8 = 0\n(x - 8)(x - 1) = 0\nx = 8 or 1\ny^(1/3) = 8 or 1\ny = 2 or 1
858799
This depends on the specific Lie group. The exponential map for GL^+(n,R) is not surjective (for n=2, [[-1 0][0 -2]] is not an exponential), but is is surjective for GL(n,C). Notice that it cannot be surjective unless the group is connected [GL(n,R) is not]. The map tends not to be injective. For example exp[[0 0][0 0]] and exp[[2pi*i 0][0 0]] are both the identity. Compactness considerations give negative answers for bijectivity for all compact Lie groups.\n\nThe exponential map is bijective on a neighborhood of the identity for every Lie group. The reason is that the differential map is essentially the identity and so the Inverse Function Theorem applies. Cleartly it is both injective and surjective for R^n.
559200
By order of precedence of operations:\n\ny = 2^1/2  is the same as\ny = (2^1)/2  which is just\ny = 2/2 which yields\n\ny = 1 <<<<<<<<<<<<<<<<<<<<<<
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]
70580
Well, I don't really understand you. This is the best I could.\n\nI think it is true. For example, the cube root of -8, which is x, is -2, which is y. y^3 is -8, which is x. So, yes, I think it's true.