What is differance between Derivative and integration?
609321
One is the inverse of the other
664811
Well,... yes. Your question is flawed though; if the derivative is continuous at a point (say, at x=a), then the derivative must exist at x=a (by definition of continuity). But then, of course, the function is differentiable at x=a since this simply means that the derivative exists at x=a
21931
It's like the difference between an exponent and a polynomial.\n\nDifferential equations are equations that can contain multiple derivatives (first derivative, second derivative, etc.), each with a coefficient. \n\nhttp://opera.answers.com/Differential%20Equations
452294
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
1068054
well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
586082
integrate by parts.\n\nintegral(lnx dx)\n= (lnx)(x) - integral (x d(lnx))\n= xlnx - integral (x (1/x) dx)\n= x lnx - integral (1 dx)\n= x lnx - x + c
840388
This is differentiation of the composite function \n\nJ(x) = I(a(x),b(x))f(t)dt\n\nwhere \nI(a,b) is the integral from a to b, \na(x) = tan(x), b(x) = x^2 \nand f(t) = sqrt(1+t^4).\n\nApply the Chain Rule and then the Fund. Thm of Calc., which implies that dI/da = -f(a) and dI/db = f(b). These should be written with curly d's because they are partial derivatives of I (as a fctn of the two integration limits). And work out da/dx and db/dx. So \n\ndJ/dx = (dI/da) (da/dx) +  (dI/db) (db/dx)\n = -f(a(x)) (da(x)/dx) + f(b(x)) (db(x)/dx)\n\nNow just put in your particular functions f, a and b.
376225
differentiation is the process of finding the rate of change of a gradient.\n( ^= to the power)\n\nsay you are given an equation y=x^n\ndo differentiate, take the power of x, and multiply it by its coefficient. in this case, x has a coefficient of 1. \ny=nx^n\nthe next step it to subtract 1 from the base of x (its power)\ny=nx^n-1\ndifferentation is now complete\n\ne.g.\ny=3x^4\ndy=12x^3\nno matter what the power of x is, just multiply it by the coefficient of x, and subtract 1 from it. (do not subtract then multiply)\n\nIntergration is the opposite, and it too complicated fot this post. talk to you teacher, of go to www.s-cool.co.uk
779428
I haven't got a clue what you are talking about, but the fundamental theorem of Calculus is that differentiation and integration are the reverse process of each other.\n\nThis link between differentiation and integration is what Newton and Leibniz discovered independently in the late 1600's.  There is a common mis-conception that Newton and Leibniz invented calculus, but calculus dates right back to the Greeks and Eudoxus and his method of exhaustion is the earliest form of integration.\n\nI think you need to re-word your question if you want some productive help.
1159338
Integrate using the method of substitution based on the fact that the derivative of log(y) + C = 1/y\n\nlet u = log(y) then du/dy = 1/y. So du = (1/y) dy\n\nUsing these substitutions we have:\n\nInt 1/y*log(y) dy  = Int 1/log(y)*(1/y) dy\n\n= Int 1/u du  = log(u)+C = log(logy)+C, since u = log(y)\n\nThis is correct since from the chain rule:\n\n d/dy log(logy)+C= 1/log(y)*d(logy)/dy= 1/(y*logy)
1121843
the given function is not differentiable at x=0\n\ntherefore,   f(x) = x   for x>0        dy/dx = 1\n                           f(x) = -x  for x<0        dy/dx = -1
79903
Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
1015799
Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
787409
that would be if it was an indefinite integral\nbut in the case of definite one, it would be the difference of the values of the function at the limits of the integral.
761317
We have to find derivative of, y = the integral where lower limit is e^x and upper limit is 0 of sin^3 t dt.\n\ny = int(e^x)(0) [sin^3 t] dt     [I would write lower limit in first bracket and upper limit in bracket after that, so dont confuse]\n\nNow, the derivative of the integrand f(t)=sin^3 t is f'(t)=3 (sin^2 t) *cos t\ndy /dt= d {int(e^x)(0) [sin^3 t] dt }/dt\n=>dy /dt= d {f(t)}/dt (at 0) - d {f(t)}/dt (at e^x)\n=>dy /dt= f'(t) (at 0) - f(t) (at e^x)\n=>dy /dt= [3 (sin^2 t) *cos t] (at 0) - [3 (sin^2 t) *cos t ] (at e^x)\n=>dy /dt= [3 (sin^2 0) *cos 0] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= [3 (0) *1] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= -[3 (sin^2 (e^x)) *cos (e^x) ]\n\nI hope that answers your question.\n\nAll the bset
480262
the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated