Let z = x +iy. How would you compute the derivitive of f(z)? I know f(z) could be anything, but please give some examples.
214541
A complex function is differentiable at a point z0 if the limit z -> z0  (f(z) - f(z0))/(z - z0) exists. However, the differivative no longer has any geometrical interpretation.\n\nThe differentiation can be done in a fashion that resembles real differentiation using the definition:\n\nlim dz -> z   ( f(z0 + dz) - f(z0) ) / dz\n\nHowever since z is a complex plane, the limit as dz -> z can approach from any direction. If it is differentiable, the value of the derivative is the same when approached from any direction, so a path is chosen that makes the calculations easy, typically keeping the real or imaginary part constant and just approaching in a direction parallel to either axis. The effect is that the limit calculation is done on a single variable. But this only finds the derivative at a single point.\n\nEquations for the derivative of a function can be found by using the Cauchy-Riemann Equations, where valid, if \nw = f(z) = u(x,y) + i v(x,y)\n\nin which case if partial du/dx = dv/dy AND partial dv/dx = -du/dy AND all theses derivatives are continuous, then \n\ndw/dz = partial du/dx + i dv/dx = partial dv/dy - i du/dy
536077
Hint:\nPick y and z different. Let \nf(x)=d(x,y)-d(x,z).\nThen f(y)<0 and f(z)>0. Show that the range of f is uncountable.
556109
nested function\nf(g(x))=f'(x)g'(x)\ng(x) = 7x\nf(x) = ln [7x]\n\nf'(x) = 7\ng'(x) = 1/x\n\nd(ln 7x)/dx = 7/x\n\nNow that I think about it, I'm not sure I like that answer either.  Maybe it's 7/7x or 1/x.
448079
You don't subtract one from the exponent on e.\nIn fact, the exponent on e NEVER changes when you differentiate or integrate (Remember that and you should be fine).\nYour answer is close,\ny*e^(x*y)cos(x*y^3*z^ 2 ) - sin(x*y^3*z^2 )*y^3*z^2\nAlso, is y or z a function of x?  If it is, then you have to find the derivative implicitly.\nAn example involving e.\nFind the deriviative of e^(x^5 + 3x^2 - x + 5)\nThe answer is\n(x^5 +3x^2 -x +5)' e^(x^5 +3x^2 -x +5)\n= (5x^4 +6x -1) e^(x^5 +3x^2 -x +5)
427545
Integrating the derivitive leaves you with the initial equation x^2 since an integral is the opposite to a derivitive much like addition is the opposite of subtraction.  If you add 2 then subtract 2 you have the same answer.  As for the area under the curve, this will indeed be your entire population.
594125
I think you're referring to synthetic division - maybe this wasn't clear to the previous responders.\n\nThe remainder when f(z) is divided by z^2+1 is some first degree polynomial in z, az+b:\n\nf(z) = Q(z)(z^2+1) + (az + b)\n\nSince z^2=(z+i)(z-i), the first term is divisible by both, and the remainders for division by (z-i) and (z+i) respectively are the same as the remainders of az+b divided by z-i and z+i:\n\naz+b = a(z-i) + (b+ia) --> b+ia=1-i\naz+b = a(z+i) + (b-ia) --> b-ia=1+i\n-->\nb = 1 , a = -1\n\nIn other words, the answer is:\n\n-z + 1
406325
The derivative is [(x-4)- (x+1)]/(x-4)^2 = -5/(x-4)^2\n\nIf we use the definition of the derivative:\n\n\ndy/dx = Lim h -> 0 {f(x+h) -f(x)}/h\n= lim h-> 0 [(x+h+1)/(x+h-4) - (x+1)/(x-4)]/h\n                   (x+h+1)(x-4)-(x+1)(x+h-4)\n= lim x-> 0 ---------------------------------------\n                   (x+h-4)(x-4)h\n\n= lim h-> 0 -5h/(x+h-4)(x-4)h = -5/(x-4)^2
391701
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
208375
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards
486143
well...both answers that were given arnt wrong...but not entirely right\n\nok...so you are given some equation you find its derivitive\n\nim going to assume you now have solved for dy/dx ...in other words you have an equation\n\ndy/dx = mess of an equation...most likely a fraction\n\nok, so now you need the second derivitive...if your dy/dx has some y's involved on the right of the equal sign your probably panicking...dont\n\ntake your derivitive..probably involving the quotient rule, and when you take derivitives of things with a y in them remember to put the dy/dx\n\nso your final equation will be\n\nd2y/dx2 = some x terms, some xy terms...maybe a fraction and probably some stuff with dy/dx's attached\n\nand you say...now what do i do...well you already know what dy/dx is..you solved for it initially...its the first derivative with respect to x\n\nyou now can substitute your dy/dx equation anywhere you see a dy/dx...and some cancelling will ensue...and life will be good\n\nhope this helped\n\nmatttlocke
472298
You differentiated and found it has relative min at x=1/3?  but if you differentiate again you find second deriv is\n54x - 1/(x^2), and I think this goes from \nnegative when x<1/cuberoot(54) to positive when it's greater, so that means it has an inflection.  Also means it's concave down when \nx<1/cuberoot(54), so the answer must be 5, not 2.  Unless I made a mistake!\n\nPS  Dave is wrong about 1.  Domain is all positive real numbers, because lnx is undefined for 0 and neg.
464653
(2x-1)/x= 2 - 1/x\n\nto show it is bijective, you need to show that it is injective and surjective:\ninjective: f(x)=f(y) => 2 - 1/x = 2 - 1/y\n=> - 1/x = -1/y => 1/x =1/y => x=y, so f(x) is injective\nsurjective: let y be in IR -{2}, then\ny= 2-1/x if\n1/x = 2-y\nso x= 1/(2-y) therefore\nf( 1/2-y) =y and f(x) is surjective.\n\nnow the inverse is:\nf^{-1} (y) =  1/(2-y).\n\nf( [-1,2]-0 ) =all reals except 0\nactually the codomain is not correct, it should not include the 0\n\ns
372037
You will have to do an epsilon-delta proof.\nNotice that\n|f(x)-f(y)|=\n|x^3-y^3 -x+y|=\n|x-y|*|x^2+xy+y^2 -1|<=\n|x-y|*[|x|^2 +|x|*|y|+|y|^2 +1]\n<=148*|x-y|
559969
The quick answer is to take the derivative of f(x) ...\n\nf'(x) = 2 (2x - 4) 2 = 4 (2x - 4)\n\n... and evaluate it at x = 4\nto get a slope for the tangent line of\nf'(4) = 16.  [Ask if you need to know\nwhy / how to do this.]\n\nNow you want the equation of a line through (4, 16) with slope 16.  That would be\n\n(y - 16) = 16 (x - 4)\n\nor\n\ny = 16 x - 48\n\nCheck: it has slope 16 and passes through (4, 16).
13129
F'(x) is the derivitive of F(x), which means it describes the change in F with respect to x.  Another way to say this is that F'(x) is the slope of F(x).  \n\nTo sketch the graph of F'(x), look at your graph of F(x).  Pick a point on that graph (maybe the first point). See what the slope (remember rise over run?) of F(x) is at that point.  That slope is the derivitive F' for that value of x.  Plot this on your F'(x) skecth.  Go the the next point and repeat.
418375
If h(x) is I-able, then h(x) is defined and exists for any x in between [a,b]. So Int{x=a until b} of h(x)*dx=H(b)-H(a). And back: If h(x) is defined in [a,b], then h(x) is I-able, being |H(b)-H(a)| a definite number.\nNow if both f(x) & g(x) are defined, then h(x)=f(x)*g(x) is also defined, thus h(x) is I-able.\nIf g(x)=f(x), then f^2(x) is also I-able.\nmay it help you Kevin!
648581
Okay this is one of those problem where you are kinda working your way backwards so here we go:\n\nFirst recall that a point of inflection is where a fuctions second derivative equals zero, in our case we are told this happens at x=3. Okay so first lets find the second derivative:\n\nIm gonna rewrite  f(x) =square root{x+1} + b/x\nas f(x)=(x+1)^(1/2) +bx^(-1)               recall that a square root is a power of 1/2 and having an x in the denominator is having it eing raised to a negative 1\n\nOkay so f'(x)=1/2(x+1)^(-1/2) - bx^(-2)\n\nand f''(x)=-1/4(x+1)^(-3/2) + 4bx^(-3)\n\nNow using the fact that f(x) has an inflection point at x=3, meaning that f''(x)=0 when x=3, we can substitute those two things into the above equation. Here we go:\n\nf''(x)=-1/4(x+1)^(-3/2) + 4bx^(-3)\n0=-1/4(3+1)^(-3/2) + 4b(3)^(-3)      Now simplify to solve for b\n0=-1/4(4)^(-3/2) + 4b(1/27)\n0=-1/4(1/8) +4/27b\n0=-1/32 + 4/27b\n\n1/32=4/27b\n1/32(27/4)=4/27(27/4)b\n27/128=b    \n\nThere it is, if my multiplication and stuff doesnt fail me b=27/128.\n\nHope that helped!