What is the derivitive of x^2?
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
517319
Using the definition of the derivative.\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n\nLet f(x) = mx + b\nf(x+h) = m(x+h) + b\n\nf(x+h) - f(x)\n= m(x+h) + b - (mx+b)\n= mx + mh + b - mx - b\n= mh\n\n[f(x+h)-f(x)]/h \n= mh/h\n= m\nTherefore,\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n= lim(as h->0) of m\n= m
3246
So for the first derivative:\n\n4(x-1)\n-----------\n3x^(2/3)\n\nThe second derivative:\n\n4(x+2)\n-----------\n9x^(5/3)
746686
well, here goes nothing.  i have an answer, but i'm not 100% sure if it's right.  as you said, (V0/g) is a constant.  the derivitive of sin(2x) is 2cos(2x).  set equal to 0 and i got:\n(V0/g)2cos(2x)=0\ndivide both sides of equation by g/2V0:\ncos(2x)=0\nthen i graphed the function of cos(x) and saw that it equalled 0 at (pi/2) and (3pi/2).  we're looking for minimum so we keep (pi/2) and discard (3pi/2).  since we're dealing with 2x and not x we have:\n2x=(pi/2)\ndivide both sides by 2\nx=(pi/4) or 45 degrees.  hope this matches up to the back of the book
972569
By definition of the derivative\ng'(x) = lim h=>0 [g(x + h) - g(x)] /h =  lim h=>0 [3xh + h^2]/h = \nlim h=>0 [3x + h] = 3x
1181156
It looks like that first 1 should be an i. In this case,\ni^z=exp(i*pi/2*z),\nwhere I am using the principal branch of the logarithm to define the exponential.\nThen the derivative will be \n(i*pi/2)*exp(i*pi/2*z)=\npi/2*i^(z+1)\n\nEasy.
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]
213471
x^2 is not equivalent to x added to itself x times. This is only true for natural numbers, and the derivative as you took it is not correct for natural numbers, as the function f(x) = x+x+... (x times) is only defined on N, while you are comparing it to the function g(x) defined on R.
866864
Just differentiate it. You need f'(x) < 0.\n\nBy the quotient rule, f'(x) = (2x(x^2 + 4) - (x^2)(2x))/(x^2+4)^2.\nNow, since the bottom is a square, thus positive, you just need the numerator to be < 0:\n2x(x^2 + 4) - x^2 * 2x < 0\n2x^3 + 8x - 2x^3 < 0\n8x < 0\nx < 0\nThus it is decreasing on (-∞,0).
179824
Its preety easy. The answer is (3X^2)(3^X). Just take the derivative of X^3. Don't do any thing with 3^X cuz X is an exponent and when a X is an exponent you just leave it like that.
831528
"What i know: it can also be written as (x^2 + cos x) ^ -1/2\nand the derivitive of that = (2x - sin x) ^ -1/2"\n\nWhat you said here was to take the derivative of the inside function, but what you do is take the derivative of the outside function and multiply it by the derivative of the inside function.\n\nIf you have a composition of two functions such as h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x) [or as in your notes, f(g) * g']\n\nThat is why you must use the chain rule for 1 and 3 but not for 2, since 2 does not involve a composition:\n\n1. f(x) = (x^2 + cos x) ^ (-1/2)\n\nf(x) = g(h(x)) where g(x) = x^ -1/2 and h(x) = x^2 + cos x\n\n3.  f(x) = 2 ^ (x^2 - sin x)\n\nThis isn't a product- it's an exponential function of another function.\n\nf(x) = g(h(x)) where g(x) = 2 ^ x and h(x) = x^2 - sin x
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
100694
f= {square root(y)}  with\ny = 460x - x^2  in which ^2 = squaire\nf'= 1/[2f) and y'= 460-2x\nI use the chain rule. (That's the name I learnt in Europe)\nThe wanted derivate to x is f ' y' =\n{(460-2x)/2}/{square root(460x - x^2)}.
572493
=x(x^3-2)\n--> x=0 or x=2^(1/3) or -2^(1/3)
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.
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.
764509
If you mean (x^4 - 2)/x, isn't it easier to just write it like that, instead of trying to draw the fraction bar?\n\nSplit it into two fractions: x^4/x - 2/x.\nThe derivative of x^4/x=x^3 is 3x^2.\nThe derivative of -2/x is 2/x^2.\n\nThe final answer is 3x^2+2/x^2 or (3x^4 + 2)/x^2.\n\nAlternatively, you can use the quotient rule.\nf= x^4-2 and g=x\nThe derivative of f/g is (f'g - g'f)/g^2.\nf'=4x^3\ng'=1\nf'g=(4x^3)(x)=4x^4\ng'f=(1)(x^4-2)=x^4-2\nf'g-g'f= 4x^4-(x^4-2)=4x^4-x^4+2=3x^4+2\nThe final answer (as above) is (3x^4+2)/x^2.