{"ret_code": 0, "ret_str": "Successful", "tot_pages": 40, "hits": [{"docid": 88726, "score": 4.289, "title": "Prove that there exists a choice of [imath]\\pm[\/imath] signs", "url": "http://math.stackexchange.com/questions/1187961/prove-that-there-exists-a-choice-of-pm-signs", "snippet": "\\pi\\le y\\le x\\le \\pi[\/imath]. (If [imath]xx<\/em>+\\cos y<\/em>+\\cos (x<\/em>-y<\/em>)+1\\\\ &= 2\\cos\\frac<\/em>{x<\/em>+y<\/em>}{2}\\cos\\frac<\/em>{x<\/em>-y<\/em>}{2}+2\\cos^2\\frac<\/em>{x<\/em>-y<\/em>}{2}\\\\ &= 2\\cos\\frac<\/em>{x<\/em>-y<\/em>}{2}\\left(\\cos\\frac<\/em>{x<\/em>+y<\/em>}{2}+\\cos\\frac<\/em>{x<\/em>-y<\/em>}{2}\\right)\\\\ &= 4\\cos\\frac<\/em>{x-y}{2}\\cos\\frac<\/em>{x}{2}\\cos\\frac{y}{2},\\\\ B&=-\\cos x-\\cos y+\\cos (x-y)+1\\\\ &= -2\\cos\\frac{x+y}{2}\\cos\\frac{x-y}{2}+2\\cos^2\\frac{x-y}{2}\\\\ &= 2\\cos\\frac{x-y}{2}\\left(-\\cos\\frac{x+y}{2}+\\cos\\frac{x-y}{2}\\right)\\\\ &= 4\\cos\\frac{x-y}{2}\\sin\\frac{x}{2}\\sin\\frac{y}{2},\\\\ C&=\\cos x-\\cos y-\\cos (x-y)+1\\\\ &= 2\\sin\\frac{x+y}{2}\\sin\\frac ... "}, {"docid": 165375, "score": 4.258, "title": "Solve in integers the equation [imath]\\left\\lfloor\\frac{x^2-y^3}{x+y^2} \\right\\rfloor=1+x-y[\/imath]", "url": "http://math.stackexchange.com/questions/1837842/solve-in-integers-the-equation-left-lfloor-fracx2-y3xy2-right-rfloor", "snippet": "^3}{x+y^2}-1-x+y<1[\/imath] 2) [imath]\\lfloor\\frac{x^2-y^3}{x+y^2} \\rfloor=\\lfloor\\frac{x^2+2xy^2+y^4-2xy^2-y^4-y^3}{x+y^2} \\rfloor=x+y^2-\\lfloor\\frac{2xy^2+y^4+y^3}{x+y^2} \\rfloor[\/imath] So the challenge seems to be to show that we must have [imath]x=-1[\/imath] or [imath]x=-2[\/imath]. We have \\begin{align} \\lfloor \\frac<\/em>{x<\/em>^2 - y<\/em>^3}{x<\/em> + y<\/em>^2} \\rfloor &= \\lfloor \\frac<\/em>{x<\/em>^2 + xy^2 - xy - y<\/em>^3}{x<\/em> + y<\/em>^2} - \\frac<\/em>{xy^2 - xy}{x<\/em> + y<\/em>^2}\\rfloor \\\\ &= \\lfloor \\frac<\/em>{(x<\/em>-y<\/em>)(x<\/em> + y<\/em>^2)}{x<\/em> + y<\/em>^2} - \\frac<\/em>{xy^2 - xy}{x + y^2}\\rfloor \\\\ &= \\lfloor x - y - \\frac<\/em>{xy^2 - xy}{x + y^2}\\rfloor \\\\ &= x - y + \\lfloor - \\frac<\/em>{xy^2 - xy}{x + y^2}\\rfloor \\end{align} Therefore, we further have [imath] \\lfloor - \\frac{xy^2 - xy}{x + y^2}\\rfloor = 1 [\/imath] or say [imath] 1 \\leq - \\frac{xy^2 - xy}{x + y^2} < 2 \\tag{1} [\/imath] Observe that [imath]x[\/imath] can not be non-negative because otherwise [imath]- \\frac{xy^2 - xy}{x + y^2}[\/imath] wo ... 2y^2}{y^2 - y} \\leq 4 \\quad\\text{(note that [\/imath][imath] is an integer)} [\/imath][imath] Finally, we know that [\/imath][imath] x \\in \\{-1, -2, -3\\} [\/imath][imath] and enumerating possible [\/imath]x[imath] can solve the problem. Why [\/imath][imath](-x)<\\frac{2y^2}{y^2-y}\\le 4?[\/imath][imath] @Roman83 [\/imath]-x < \\frac<\/em>{2y^2}{y^2 - y}[imath] because of inequality (2). To see that [\/imath]\\frac{2y^2}{y^2 - y} = \\frac{2}{1 - \\frac{1}{y}} \\leq 4[imath], observe that [\/imath]\\frac{2}{1 - \\frac{1}{y}}[imath] is maximized when [\/imath]y = 2[imath]. If [\/imath]x\\ge0[imath] it is easy to show that the equation cannot hold. If [\/imath]x=y=0 ... "}, {"docid": 130192, "score": 4.255, "title": "Solve a system of equations.", "url": "http://math.stackexchange.com/questions/859989/solve-a-system-of-equations", "snippet": "Solve a system of equations. I have a system of equations: \\begin{align} & x<\/em>_{21} (\\frac<\/em>{\\partial}{\\partial x<\/em>_{11}}f_{1111})( x<\/em>_{11} , x<\/em>_{21}, y<\/em>_{11} , y<\/em>_{21} ) + \\frac<\/em>{y<\/em>_{21}}{x<\/em>_{11}^2} (\\frac<\/em>{\\partial}{\\partial y<\/em>_{11}}f_{1111})( x<\/em>_{11} , x<\/em>_{21}, y<\/em>_{11} , y<\/em>_{21} ) \\\\ & = f_{2111}(x<\/em>_{11}, x_{21}, y<\/em>_{11}, y<\/em>_{21}) + x_{11}^{-1}x_{22} f_{1121}( x_{11}, x_{21}, y_{11}, y_{21} ) - \\frac<\/em>{1}{2} x_{22} y_{21}. \\end{align} \\begin{align} & x_{21} (\\frac<\/em>{\\partial}{\\partial x_{11}}f_{1121})( x_{11} , x_{21}, y_{11} , y_{21} ) + \\frac<\/em>{y_{21}}{x_{11}^2} (\\frac<\/em>{\\partial}{\\partial y_{11}}f_{1121})( x_{11} , x_{21}, y_{11} , y_{21} ) \\\\ & = f_{2121}(x_{11}, x_{21}, y_{11}, y_{21}). \\end{align} \\begin{align} & x_{21} (\\frac<\/em>{\\partial}{\\partial x_{11}}f_{ 2121})( x_{11} , x_{21}, y_{11} , y_{21} ) + \\frac{y_{21}}{x_{11}^2} (\\frac{\\partial}{\\partial y_{11}}f_{2121})( x_{11} , x_{21}, y_{11} , y_{21} ) = 0. \\end{align} \\begin{align} & x_{21} (\\frac{\\partial}{\\partial x_{11}}f_{ 2111})( x_{11} , x_{21}, y_{11} , y_{21} ) + \\fr ... "}, {"docid": 21881, "score": 4.248, "title": "Finding partial derivative plug in a value?", "url": "http://math.stackexchange.com/questions/703583/finding-partial-derivative-plug-in-a-value", "snippet": " [imath]\\frac{d}{dy}=x^2\\frac{1}{1+(x\/y)^2}(-x\/y^2)=\\frac{-x^3}{y^2+x^2}[\/imath] so for [imath](-1,1)[\/imath] I get [imath]1\/2[\/imath] Also, you have not evaluated the [imath]x[\/imath] derivative properly. [imath]\\arctan(-1)[\/imath] is not [imath]\\frac{\\pi}{2}[\/imath]. For [imath]F_x[\/imath], we have \\begin{align} F_x<\/em> &= \\frac<\/em>{\\partial}{\\partial x<\/em>} (x<\/em>^2 \\tan^{-1}\\frac<\/em>{x<\/em>}{y<\/em>})\\\\ &= \\frac<\/em>{\\partial}{\\partial x<\/em>} (x<\/em>^2) \\cdot \\tan^{-1} \\frac<\/em>{x<\/em>}{y<\/em>} + x<\/em>^2 \\cdot \\frac<\/em>{\\partial}{\\partial x} \\tan^{-1} \\frac<\/em>{x}{y<\/em>} \\\\ &= 2x \\tan^{-1} \\frac<\/em>{x}{y<\/em>} + \\frac<\/em>{x^2}{y<\/em>} \\frac{1}{1+(\\frac{x}{y<\/em>})^2} \\end{align} So \\begin{align} F_x(-1,1) &= 2 \\cdot (-1) \\cdot \\tan^{-1} (\\frac{-1}{1}) + \\frac{(-1)^2}{1} \\frac{1}{1 + (\\frac{-1}{1})^2} \\\\ &= 2 \\cdot (-1) \\cdot -\\frac{\\pi}{4} + \\frac{1}{2} \\\\ &= \\frac{\\pi + 1}{2} \\end{align} For [imath]F_y[\/imath], we have \\begin{align} F_y<\/em> &= \\frac{\\partial}{\\partial y<\/em>} (x^2 \\tan^{-1}\\frac{x}{y}) \\\\ &= x^2 \\frac{\\partial}{\\partial y} \\tan^{-1}\\frac{x}{y} \\\\ &= x^2 \\frac{1}{1+(\\frac{x}{y})^2} \\frac{\\partial}{\\partial y} (\\frac{x}{y}) \\\\ &= x^2 \\frac{1}{1+(\\frac{x}{y})^2} (-\\frac{x}{y^2}) \\\\ &= \\frac{-x^3 y ^2}{1+(\\frac{x}{y})^2} \\end{align} So \\begin{align} F_y(-1,1) &= \\frac{-(-1 ... "}, {"docid": 103404, "score": 4.247, "title": "Prove that [imath]\\frac{1}{x(1-y)} +\\frac{1}{y(1-z)} +\\frac{1}{z(1-x)} \\ge \\frac{3}{xyz+(1-x)(1-y)(1-z)} [\/imath]", "url": "http://math.stackexchange.com/questions/1018431/prove-that-frac1x1-y-frac1y1-z-frac1z1-x-ge-frac3xy", "snippet": "[imath]\\frac{1}{x(1-y)} +\\frac{1}{y(1-z)} +\\frac{1}{z(1-x)} \\ge \\frac{3}{xyz+(1-x)(1-y)(1-z)} [\/imath] Let [imath]x,y,z[\/imath] be real numbers in the range of [imath](0,1)[\/imath]. Prove that [imath]\\frac{1}{x(1-y)} +\\frac{1}{y(1-z)} +\\frac{1}{z(1-x)} \\ge \\frac{3}{xyz+(1-x)(1-y)(1-z)}.[\/imath] \\begin{eqnarray} &&[\\frac<\/em>{1}{x<\/em>(1-y<\/em>)} +\\frac<\/em>{1}{y<\/em>(1-z)} +\\frac<\/em>{1}{z(1-x<\/em>)}][xyz+(1-x<\/em>)(1-y<\/em>)(1-z)]\\\\ &=&[\\frac<\/em>{yz}{1-y<\/em>}+\\frac<\/em>{(1-x<\/em>)(1-z)}{x<\/em>}]+[\\frac<\/em>{xz}{1-z}+\\frac<\/em>{(1-x<\/em>)(1-y<\/em>)}{y<\/em>}]+[\\frac<\/em>{xy}{1-x<\/em>}+\\frac{(1-y<\/em>)(1-z)}{z}]\\\\ &=&[\\frac{z}{1-y<\/em>}+\\frac{1-z}{x<\/em>}-1]+[\\frac{x}{1-z}+\\frac{1-x}{y}-1]+[\\frac{y}{1-x}+\\frac{1-y}{z}-1]\\\\ &\\geq& 6\\sqrt[6]{\\frac{z}{1-y}\\cdot\\frac{1-z}{x}\\cdot\\frac{x}{1-z}\\cdot\\frac{1-x}{y}\\cdot\\frac{y}{1-x}\\cdot\\frac{1-y}{z}}-3=3 \\end{eqnarray} It's obviously true after following substitution and full expanding. [imath]x=\\frac{a}{a+1}[\/imath], [ima ... "}, {"docid": 194331, "score": 4.229, "title": "A transformation formula for the digamma function", "url": "http://math.stackexchange.com/questions/2824552/a-transformation-formula-for-the-digamma-function", "snippet": "A transformation formula for the digamma function I derived an astonishing relationship that includes the digamma function:\\begin{align}\\upsilon(x<\/em>,a)&=\\gamma+\\psi(1-aix)-\\frac<\/em>{a\\pi}2\\operatorname{csch}^2a\\pi x<\/em>+\\frac<\/em>{i\\pi}{e^{2a\\pi x<\/em>}-1}-\\frac<\/em>1{x<\/em>(e^{2a\\pi x<\/em>}-1)}\\\\[2ex] &\\quad+\\sum_{k\\ge1}\\frac<\/em> xk\\bigg(\\frac<\/em>1{(x<\/em>+k)(e^{2a\\pi(x<\/em>+k)}-1)}-\\frac<\/em>1{(k-x<\/em>)(e^{2a\\pi(k-x)}-1)}\\bigg)\\\\[4ex] &=-\\gamma-\\psi(1-x)-\\frac<\/em>\\pi{2a}\\csc^2\\pi x+\\frac<\/em>{i\\pi}{e^{2i\\pi x}-1}-\\frac i{ax(e^{2i\\pi x}-1)}\\\\[2ex] &\\quad-ai\\sum_{k\\ge1}\\frac xk\\bigg(\\frac1{(aix+k)(e^{2k\\pi\/a+2i\\pi x}-1)}-\\frac1{(k-aix)(e^{2k\\pi\/a-2i\\pi x}-1)}\\bigg) \\end{align}This formulation looks of the modular type, and reminds me of Ramanujan's series reformulations. Can anyone confirm if he had derived ... the answer contains at least five pages' worth of calculations. I'll elaborate here on how I got my derivation: the formula[imath]\\frac1{e^{2a\\pi x}-1}=\\frac1{2a\\pi x}-\\frac12+\\frac1\\pi\\sum_{k\\ge1}\\frac{ax}{k^2+a^2x^2}[\/imath]allows me to re-evaluate [imath]S(x,y,a)[\/imath] as\\begin{align}&\\sum_{k\\in\\mathbb Z}\\frac y<\/em>{(k+x)(x-y<\/em>+k)}\\bigg(\\frac1{2a\\pi(k+x)}-\\frac12+\\frac1\\pi\\sum_{n\\ge1}\\frac{a(k+x)}{n^2+a^2(k+x)^2}\\bigg)\\\\[2ex] &=\\sum_{k\\in\\mathbb Z}\\bigg(\\frac y<\/em>{2a\\pi(k+x)^2(x-y<\/em>+k)}-\\frac y<\/em>2\\cdot\\frac1{(k+x)(k+x-y<\/em>)}+\\sum_{n\\ge1}\\frac{ay\/\\pi}{(k+x-y<\/em>)(n^2+a^2(k+x)^2)}\\bigg)\\\\[2ex] &=\\frac1{2a\\pi}\\sum_{k\\in\\mathbb Z}\\bigg(\\frac1{(k+x)(k+x-y<\/em>)}-\\frac1{(k+x)^2}\\bigg)-\\frac y2\\sum_{k\\in\\mathbb Z}\\frac1{(k+x)(x-y+k)}\\\\[2ex] &\\quad+\\sum_{k\\in\\mathbb Z,\\,n\\ge1}\\frac{y\/(a\\pi)}{2in\/a(k+x-y)}\\bigg(\\frac1{k+x-in\/a}-\\frac1{k+x+in\/a}\\bigg)\\\\[2ex] &=\\bigg(\\frac1{2a\\pi}-\\frac y2\\bigg)\\sum_{k\\in\\mathbb Z}\\frac1{(k+x)(k+x-y)}-\\frac\\pi{2a}\\csc^2\\pi x\\\\[2ex] &\\quad+\\frac ... "}, {"docid": 73553, "score": 4.214, "title": "Finding the nth derivative of functions, in particular y=tan(x)", "url": "http://math.stackexchange.com/questions/2314942/finding-the-nth-derivative-of-functions-in-particular-y-tanx", "snippet": "Finding the nth derivative of functions, in particular y<\/em>=tan(x<\/em>) Since [imath]\\frac{d y}{d x} = -2 \\sin2x[\/imath], and [imath]\\frac{d^2 y}{d x^2} = -4 \\cos2x[\/imath], then [imath]\\frac{d^2 y}{d x^2} = -4y[\/imath], so [imath]\\frac{d^3 y}{d x^3} = -4\\frac{d y}{d x}[\/imath], [imath]\\frac{d^4 y}{d x^4} = -4\\frac{d^2 y}{d x^2}[\/imath], etc. Therefore, the nth differential of [ima ... ow \\frac{d y}{d x} = x^{-1} \\Rightarrow \\frac{d^2 y}{d x^2} = -x^{-2}[\/imath], etc. So for [imath]y=\\ln(x)[\/imath], [imath]\\frac{d^n y}{d x^n} = (n-1)!(-1)^{n-1}x^{-n}, n\\geq 2[\/imath] So the question is, does an equation such as that exist for [imath]y=\\tan(x)[\/imath]? I have tried two approaches: \\begin{array}{c} \\frac<\/em>{d y<\/em>}{d x<\/em>} & = & 1+y<\/em>^2 \\\\ \\frac<\/em>{d^2 y<\/em>}{d x<\/em>^2} & = & 2y<\/em>\\frac<\/em>{d y<\/em>}{d x<\/em>} \\\\ \\frac<\/em>{d^3 y<\/em>}{d x<\/em>^3} & = & 4(\\frac<\/em>{d y<\/em>}{d x<\/em>})^2 & - & 2\\frac<\/em>{d y}{d x<\/em>} \\\\ \\frac<\/em>{d^4 y}{d x<\/em>^4} & = & 20y(\\frac<\/em>{d y}{d x})^2 & - & 4y\\frac{d y}{d x} \\\\ \\frac{d^5 y}{d x^5} & = & 20(\\frac{d y}{d x})^3 & + & 68(\\frac{d y}{d x})^2 & - & 72 \\frac{d y}{d x} \\\\ \\end{array} There are some patterns you can spot but it didn't seem very promising so I moved on. \\begin{array}{c} \\frac{d y}{d x} & = & 1+y^2\\\\ \\frac{d^2 y}{d x^2}& = & ... "}, {"docid": 88404, "score": 4.209, "title": "Integrate [imath]\\frac{1}{\\sqrt{x^2+y^2+z^2}}[\/imath] in parallelepiped", "url": "http://math.stackexchange.com/questions/2537508/integrate-frac1-sqrtx2y2z2-in-parallelepiped", "snippet": " layout depends on the rendering mode. For the browser I use, out of the 5 rendering modes, only the HTML-CSS mode knows how to wrap the very long lines in this question. Sorry for inconvenience, I'll try to wrap it with align environment With thr help of this post, I've got the answer: \\begin{multline*} \\int\\limits_{x<\/em>_1}^{x<\/em>_2}\\int\\limits_{y<\/em>_1}^{y<\/em>_2}\\int\\limits_{z_1}^{z_2}\\frac<\/em>{1}{\\sqrt{x<\/em>^2+y<\/em>^2+z^2}} dx dy dz = \\\\ = \\frac<\/em>{1}{2}\\Bigg[z_1^2 \\arctan\\left(\\frac<\/em>{x<\/em>_1 y<\/em>_1}{z_1 \\sqrt{x<\/em>_1^2+y<\/em>_1^2+z_1^2}}\\right)+y<\/em>_1^2 \\arctan\\left(\\frac<\/em>{x<\/em>_1 z_1}{y<\/em>_1 \\sqrt{x<\/em>_1^2+y<\/em>_1^2+z_1^2}}\\right)+x<\/em>_1^2 \\arctan\\left(\\frac<\/em>{y_1 z_1}{x_1 \\sqrt{x_1^2+y_1^2+z_1^2}}\\right)- \\\\ -z_1^2 \\arctan\\left(\\frac<\/em>{x_2 y_1}{z_1 \\sqrt{x_2^2+y_1^2+z_1^2}}\\right)-y_1^2 \\arctan\\left(\\frac<\/em>{x_2 z_1}{y_1 \\sqrt{x_2^2+y_1^2+z_1^2}}\\right)-x_2^2 \\arctan\\left(\\frac<\/em>{y_1 z_1}{x_2 \\sqrt{x_2^2+y_1^2+z_1^2}}\\right)-z_1^2 \\arctan\\left(\\frac{x_1 y_2}{z_1 \\sqrt{x_1^2+y_2^2+z_1^2}}\\right)-\\\\ -y_2^2 \\arctan\\left(\\frac{x_1 z_1}{y_2 \\sqrt{x_1^2+y_2^2+z_1^2}}\\right)-x_1^2 \\arctan\\left(\\frac{y_2 z_1}{x_1 \\sqrt{x_1^2+y_2^2+z_1^2}}\\right)+z_1^2 \\arctan\\left(\\frac{x_2 y_2}{z_1 \\sqrt{x_2^2+y_2^2 ... "}, {"docid": 149218, "score": 4.208, "title": "Transform the equation [imath](x+y)\\frac{\\partial u}{\\partial x} - (x-y)\\frac{\\partial u}{\\partial y} = 0[\/imath] with [imath]s = \\ln\\sqrt{x^2+y^2}[\/imath], [imath]t = \\arctan(y\/x)[\/imath]", "url": "http://math.stackexchange.com/questions/2823001/transform-the-equation-xy-frac-partial-u-partial-x-x-y-frac-partial", "snippet": "Transform the equation [imath](x+y)\\frac{\\partial u}{\\partial x} - (x-y)\\frac{\\partial u}{\\partial y} = 0[\/imath] with [imath]s = \\ln\\sqrt{x^2+y^2}[\/imath], [imath]t = \\arctan(y\/x)[\/imath] Let [imath]u(x, y)[\/imath] be a [imath]C^1[\/imath]-smooth function. Transform the equation \\begin{equation} (x<\/em>+y<\/em>)\\frac<\/em>{\\partial u}{\\partial x<\/em>} - (x<\/em>-y<\/em>)\\frac<\/em>{\\partial u}{\\partial y<\/em>} = 0 \\end{equation} by introducing the new variables [imath]s = \\ln\\sqrt{x^2+y^2}[\/imath], [imath]t = \\arctan(y\/x)[\/imath]. Use this to write down the general solution to the equation. I've tried multiple things with no luck as I think I'm way off. I really have no idea how to rewrite the equation in terms of [imath]t[\/imath] and [imath]s[\/imath]. Would love some help, thanks. \\begin{align} \\frac<\/em>{\\partial u}{\\partial x<\/em>} &= \\frac<\/em>{\\partial u}{\\partial s} \\frac<\/em>{\\partial s}{\\partial x<\/em>}+ \\frac<\/em>{\\partial u}{\\partial t} \\frac<\/em>{\\partial t}{\\partial x<\/em>} \\\\ &= \\frac<\/em>{x<\/em>}{x<\/em>^2+y<\/em>^2}\\frac{\\partial u}{\\partial s}- \\frac{y<\/em>}{x^2+y<\/em>^2}\\frac{\\partial u}{\\partial t} \\\\ \\frac{\\partial u}{\\partial y<\/em>} &= \\frac{\\partial u}{\\partial s} \\frac{\\partial s}{\\partial y<\/em>}+ \\frac{\\partial u}{\\partial t} \\frac{\\partial t}{\\partial y} \\\\ &= \\frac{y}{x^2+y^2}\\frac{\\partial u}{\\partial s}+ \\frac{x}{x^2+y^2}\\frac{\\partial u}{\\partial t} \\\\ (x+y)\\frac{\\partial u}{\\partial x}-(x-y)\\frac{\\partial u}{\\partial y} &= \\frac{(x+y)x-(x-y)y}{x^2+y^2} \\frac{\\partial u}{\\partial s}+ \\fr ... "}, {"docid": 166456, "score": 4.201, "title": "zero Jacobian matrix determinant and local inverse", "url": "http://math.stackexchange.com/questions/2902874/zero-jacobian-matrix-determinant-and-local-inverse", "snippet": "ave a local inverse at every point of [imath]\\mathbb R^2[\/imath]? Update: The interesting fact with this function is that we can not use the inverse function theorem to decide whether it is locally injective, since for all points [imath](x, y)[\/imath] we have [imath]Jf (x, y) = 0[\/imath]. Indeed, \\begin{align} \\det Jf(x<\/em>,y<\/em>) = & \\det \\begin{pmatrix} \\frac<\/em>{\\partial}{\\partial x<\/em>} \\frac<\/em>{ (x<\/em>^2-y<\/em>^2)}{(x<\/em>^2+y<\/em>^2)} & \\frac<\/em>{\\partial}{\\partial y<\/em>} \\frac<\/em>{ (x<\/em>^2-y<\/em>^2)}{(x<\/em>^2+y<\/em>^2)} \\\\ \\frac<\/em>{\\partial}{\\partial x<\/em>} \\frac<\/em>{ xy}{(x<\/em>^2+y<\/em>^2)} & \\frac<\/em>{\\partial}{\\partial y<\/em>} \\frac<\/em>{xy}{(x^2+y^2)} \\end{pmatrix} \\\\ =& \\det \\begin{pmatrix} \\frac{ 2x(x^2+y^2)-(x^2-y^2)2x}{(x^2+y^2)^2} & \\frac{-2y(x^2+y^2)-(x^2-y^2)2y}{(x^2+y^2)^2} \\\\ \\frac{y(x^2+y^2)-xy(2x) }{(x^2+y^2)^2} & \\frac{ x(x^2+y^2)-xy(2y) }{(x^2+y^2)^2} \\end{pmatrix} \\\\ =& \\det \\begin{pmatrix} \\frac{ 4xy^2}{(x^2+y^ ... "}]}