How do you take the derivitive of a complex function?
472993
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
347106
Okay, if you're actually serious, you REALLY should be paying more attention in class, but I'm gonna answer it for two points anyway.\n\nThe derivative of x^2 (with respect to x) is 2x.\n\nHave fun.
1125670
Let x = (-2)^(-2)^(-2)^(-2)^(-2)^(-2)^...\n\n(-2)^(-2)^(-2)^(-2)^(-2)^(-2)^... = (-2)^(-2)^(-2)^(-2)^(-2)^(-2)^...\nx = (-2)^x\nln(x) = x * ln(-2)\n\nNot a real number, that's for certain.  I have to brush up on my complex logarithmic functions.... :-(
710286
the antiderivative of 4 / (t^2 + 1)  IS \n4 arctan(t) + C`
620534
First, take the partial derivitives...\n\nSo, the partial with respect to x is...\n\n2xy^3z^6\n\nAnd the partial with respect to y is...\n\n3x^2y^2z^6\n\nAnd the partial with respect to z is...\n\n6x^2y^3z^5\n\nThis gives you the gradient...\n\n(2xy^3z^6, 3x^2y^2z^6, 6x^2y^3z^5)\n\nIn which you plug in the point...which simplifies to...\n\n(-2, 3, -6)\n\n...So this is a vector...so it should be denoted\n\n<2, 3, -6>\n\nTo find the magnitude of this vector, simply take the root of the sum of the squares...\n\nsqrt(4 + 9 + 36) which = sqrt(49) = 7\n\nCake :)
813180
three variables are related to time ...( X,Y,Z,) related to time\n\nthe total derative can be expressed by the sum of the partials... that is dx/dt + dy/dt + dz/dt = total derative....\n\nthe partial with respect to x (dx/dt)  plus\nthe partial with respect to y (dy/dt)  plus\nthe partial with respect to z (dz/dt)  equals\n\nthe total derative\n\nthe partial gives us an idea of how the function acts on that varaible
761744
use the product rule, with the first term as t and the second as (4-t)^1/2\nproduct rule says that the derivative is second term times the derivative of the first term plus first term times the derivative of the second term.  Don't forget to use the chain rule along the way.\n\nThat is\nxy would have a derivate of yx' =xy'\n\nSo here\n[(4-t)^1/2](1) + (t){[1/2(4-t)^ -1/2][-1]\n(4-t)^1/2 + -t/2(4-t)^ -1/2
940999
Differentiate the function as you would normally except that when the derivative of y is taken make sure to replace y with y'. So this would be,\n(1/2)(xy)^(-2/3)(xy') + y = 0 - y'\nThen, solve for y' by moving all the y' over to one side and factoring it out.\ny' + (1/2)(xy)^(-2/3)(xy') + y = 0\n\ny'((1/2)(xy)^(-2/3)x + 1) = -y\n\ny' = -y / ((1/2(xy)^(-2/3)x + 1)\n\nThe second derivative is (quotient rule),\ny'' = (.5(xy)^(-2/3) + 1)(-y') - (-y)(1/3)(xy)^(-5/3)xy(xy')(y)\n\ny'' = .5(xy)^(-2/3)y' - y' + (1/3)x^2*y^3*y'(xy)^(-5/3)\nFactor,\ny'' = y'{(xy)^(-5/3)[.5 - 1 + (1/3)x^2*y^3(xy)]}\n\ny'' = y'{(xy)^(-5/3)[- .5 + (1/3)x^2*y^3(xy)]}\n\n(check my work)
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)}.
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 :)
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
766972
First off, this problem involves the chain rule. I think the person who said package rule really meant product rule, which is not requried here. \n\nThe chain rule allows us to find the deriviative of a function which we don't know how to differentiate in terms of functions which we do know how to differentiate. In this case, we know how to differentiate sin(3x), but not (sin(3x))^4 (I think it is less confusing to write it that way given the lack of mathematical notation available here).  The chain rule, in general, says that, if we have two functions, g(x) and h(x), we can find the deriviative of the compostie function g(h(x)) as follows:\n(g(h(x))' = g'(h(x)) * h'(x)\nSo, let us have hour h(x) function defined as:\nh(x) = sin(3x)\nIt's derivative is then:\nh'(x) = 3cos(3x)\nNow, lets define g(x) as:\ng(x) = x^4\ng'(x) = 4x^3\ng composed on h is then:\ng(h(x)) = g(sin(3x)) = (sin(3x))^4\ng' composed on h is:\ng'(h(x)) = 4(sin(3x))^3\nNow we have everything we need to use the chain rule, and find the derivative of g(h(x))\n(g(h(x))' = 4(sin(3x))^3 * 3cos(3x) \n             = 12*cos(3x)*(sin(3x))^3\nIn practice, one doesn't need to do this process so formally. Any time you have some base function you know how to differentiate raised  to some power, to take the derivative you simply bring the power out front, lower the power by one, and mulitply by the derivative of the base.
858871
let's say its da/dx means that you are taking derivates of a respect to x.
956017
well, to differentiate a the given funtion with respect to                (1-4x^2)^1/2, we assume (1-4x^2)^1/2=y. and the function becomes sin^-1(4y) or 1/sin4y. we have to differentiate this with respect to y.\nso, the function, which is cosec4y, on differentiation gives the result as -cosec4y*cot4y*4\nsubstituting the value of y we get the final result in terms of x as\n-4cosec((1-4x^2)^1/2)*cot((1-4x^2)^1/2)\nthe long string in the cot is (1-4x^2)^1/2 ie., y
797614
The only problem in your derivation is that you are evaluating f(0, 0) as 0/0. This is not the case, the second part of the definition is that f(x, y) = 0 when (x, y) = (0, 0). Nonetheless, your result is correct: f_x(0, 0) = [h→0]lim (0/h² - 0)/h = 0. Similarly with f_y(0, 0).\n\nThis makes sense, because the restriction of f(x, y) to the x-axis is a constant function (i.e. f(x, 0) = 0 for all x), so of course the partial derivative of it w.r.t. x is constant. Similarly, if you restrict f(x, y) to the y-axis, you also get a constant function.\n\nNote that f(x, y) is not differentiable in the 2-variable sense at (0, 0) (there is no tangent plane at that point), so the multivariate chain rule will NOT hold, and the directional derivatives won't be related to the partial derivatives. This helps to highlight an important difference between the partial and total derivates - it is possible for all partial derivatives to be defined and for the function to still not be differentiable at a point.
877239
look up differentiation in your textbook