What is the anti derivative of 1/x^2????
549061
1/x^2 can be rexpressed as x^(-2). As you might know, the formula for a derivative of x^n is nx^(n-1). The antiderivative of x^n is (x^(n+1))/n+1. In this case, it would be x^-1/-1, or -x^-1. To check this, we take the derivative, which gives us (-1)(-1)x^-2, or x^-2. The answer checks out. \n\nSo, in summary, the antiderivative of x^-2 is -x^-1, with the remainder C.
942027
The first step in these type of questions is to get the expression into index form.\n\nThis becomes x - 3 x^1/2.  Also, remember x is the same as x^1\n\nSo dy/dx =  1*x^(1-1) - 3*1/2 * x^(1/2 -1)\n\n               =  1 * x^0 - 3/2 * x^(-1/2)\n\n               =  1 - 3/2* x1/sq.root x     remember x^0 =1\n\n                = 1 - 3/(2sq.root x)
745843
Use the quotient rule\nThe derivative is\n[(Q)(25000 +150 + 3Q)' - (25000 + 150 + 3Q)(Q)'] / (Q)^2\n= [(Q)(3) - (25000 + 150 + 3Q)(1)] / Q^2\n= [3Q - 25000 - 150 - 3Q] / Q^2\n= [-25000 - 150]/ Q^2\n= -25150/Q^2\nIf you want (and I would recommend this), then you can combine 25000 and 150 to get 25150.
750296
-2x + h\nIn about 5 minutes you will be taken the limit of the expression as h ---> 0.  Then, you will be told that then it will be the derivative.\n\nAfter that you take 2 - x^2 do the derivative and get -2x.
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
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
931778
Since you have conflicting answers, I'm just chiming in to say go with ManiacalAce -- he got the correct answer.\n\nCorrect second derivative is as follows:\n\n(-9/4)x^(-3/2) - (3/4)x^(-1/2)\n\n~ ♥ ~
38734
(2/3)x^3 + (5/2)x^2\nyou just take the antiderivitive of each term.  so you backtrack. instead of taking away one from the power, you add one and then divide the whole term by the new power. so the first term would be to the third power. now if you took the derivitive you would multiply by that power of three, so we need to cancel that out by dividing by three. so it's (2/3)x^3  because when you take the derivitive and that three comes down, it will be (3)(2)/(3) and the 3's will cancel out giving you 2x^2 which is what you started with. so the antiderivitive is (2/3)x^3.\n\nnow we do the second term. so we add a power, so it would be squared. now we have to cancel that power out so we'll divide the whole thing by two(or rather, multiply by a half) so the antiderivitive of 5x would be (5/2)x^2. this can be checked by finding the derivitive. multiply the (5/2) times 2 and you are left with 5 and take a power away to get 5x.  integrals are easier than you think, its just the opposite of differentiation. the general formula is:  X^n integrated is (X^n+1)/n+1 so if it's x^3 it would become x^4/4 or (1/4)x^4. again, you can check this by differentiating it. the exponent of 4 is multiplied and cancels out the fraction, and the exponent turns to a 3 and you're left with what you started with...\nhope you understand integrals now :)
705841
When you have your variable in the exponent, you can't take derivates like you would for x^n.\n\nWhen you have a constant raised to x, you take the derivative like this:\n\nd/dx ( a^x ) = a^x ln a\n\nSo, if f(x) = 2^x, then f '(x) = 2^x ln2\n\nHope that helped!\n\n.
899001
You didn't write your problem clearly.  I don't know if g(x) is:\n\n1)  x² + (7/x³) - 2, or\n2)  (x² + 7) / (x³ - 2)\n\nThe quality of the answers you receive are in direct relation to the quality of the question you write!  :-)\n\nAssuming you meant (1):\n\ng '(x) = 2x - 21x^-4\n\nAssuming you meant (2):\n\ng '(x) = (-x^4 - 21x² - 4x) / (x³ - 2)²\n\nEDITED TO ADD:\n\nHi Jeff, I got your email but couldn't respond; yahoo said your email was "unconfirmed" whatever that means!  So, I am responding to you here. Hope you see this!\n\nYOU WROTE: "I was wondering if you could explain how to work through this problem"\n\nMY RESPONSE:\n\nHi Jeff, sure I am happy to explain. Basically, when you differentiate a fraction f(x) over g(x) the rule is "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared" -- i.e., \n\n[ (g(x)*f '(x)) - (f(x)*g '(x)) ] / (g(x))^2\n\nSo in your problem f(x) = x^2 - 7\ng(x) = x^3 - 2\n\nSo derivative = [ (x^3 - 2)*(2x) - (x^2 - 7)*3x^2 ] / (x^3 - 2)^2\n\nThen just multiply the numerator and combine like terms to get:\n\n[ -x^4 - 21x^2 - 4x ] / (x^3 - 2)^2\n\nDid that help?  Let me know if you have additional questions!\n\n.
853051
We have to find derivative of, g(x) = the integral (with lower limit tanx and upper limit x^2) of 1 / [sqrt(2+t^4)] dt.\n\nNow, the derivative of the integrand f(t)=1 / [sqrt(2+t^4)] is \nf'(t)= -2*t^3 / (2+t^4)^(3/2) [By Quotient Rule]\n\nd(g(x)) /dt= d {int(tan x)(x^2) {1 / [sqrt(2+t^4)] }dt} /dt\n=>dy /dt= d {f(t)}/dt (at tan x) - d {f(t)}/dt (at x^2)\n=>dy /dt= f'(t) (at tan x) - f(t) (at x^2)\n=>dy /dt= [2*t^3 / (2+t^4)^(3/2)] (at tan x) - [2*t^3 / (2+t^4)^(3/2)] (at x^2)\n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^2)^3 / (2+(x^2)^4)^(3/2)] \n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^8) / (2+(x^16))^(3/2)]\n\nI hope that answers your question.
916561
This is a fraction, yes?\n\nThe derivative of a fraction is\n\nThe derivative of the numerator times the denominator\nminus\nThe derivative of the denominator times the numerator\nAll divided by the square of the denominator\n\nLet's do it:\n\n[-sin(x) * (1 - cos(x) - sin(x) * (1 + cos(x)] / [1 - cos(x)]^2\n[-2sin(x)] / [1 - cos(x)]^2\n\nNotice that your function is discontinuous at x = 0 and x = pi, etc., and so is the derivative.\n\nI LUV screwed up again. The derivative of cos(x) is -sin(x), not sin(x) as she has it.\n\nSteve's solution agrees with mine.
833002
Let u = x², so du = 2xdx\n\nSo, ∫ (x/1+x^4) dx =\n\n1/2 ∫ (1/1+u²) du =\n\n1/2 arctan(u) + C = \n\n1/2 arctan(x²) + C\n\n~ ♥ ~
790977
The basic formula for differentiation of x^n is n.x^(n-1).\nSo your problem can be re-written as (x-1)^1/2.\n\nHere n is 1/2\n\nTherefore n.x^(n-1) gives (1/2).(x-1)^(1/2-1) (multiplied by 1)\n\nNow (1/2-1) is -(1/2).Therefore the entire expression becomes (1/2).(x-1)^(-1/2)\n\nNow this can be written as 1/{(2(x-1)^1/2)}\n\nThat is we are bringing that -1/2 down and it becomes positive.\n\nBut anything to the power of 1/2 is the root of that 'anything'.\n\nSo your final answer should be \n\n        1\n------------------\n2(sqrt of (x-1))\n\nRemember the differention of the bracket term,that is (x-1) is 1. As d/dx(x) is one and -1 is a constant. Hence that multiplied by one in the third sentence.
460400
You would need to know how to integrate (that is, find anti-derivatives). The proceedure forthis type of problem can usually be understood after first semester calculus. The specific integrations may need some second semester calculus.
667880
a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2