150 баллов! Градиент.
дано:
функция z=arcsin(x/(x+y)) ,
точка A(5;5) ,
вектор a= -12i + 5j
Найти в точке А: grad(z) и производную по направлению вектора а

$grad(z)=frac{partial z}{partial x}bar{i}+frac{partial z}{partial y}bar{j}frac{partial z}{partial x}=frac1{sqrt{1-frac{x^2}{(x+y)^2}}}cdotleft(frac x{x+y}right)_{partial x}=frac1{sqrt{1-frac{x^2}{(x+y)^2}}}cdotfrac{x+y-x}{(x+y)^2}=frac y{(x+y)^2sqrt{1-frac{x^2}{(x+y)^2}}}==frac y{(x+y)^2sqrt{frac{x^2+2xy+y^2-x^2}{(x+y)^2}}}=frac y{(x+y)^2sqrt{frac{y(2x+y)}{(x+y)^2}}}$

$frac{partial z}{partial y}=frac1{sqrt{1-frac{x^2}{(x+y)^2}}}cdotleft(frac x{x+y}right)_{partial y}=frac1{sqrt{1-frac{x^2}{(x+y)^2}}}cdotleft(-frac x{(x+y)^2}right)==-frac x{(x+y)^2sqrt{1-frac{x^2}{(x+y)^2}}}=-frac x{(x+y)^2sqrt{frac{y(2x+y)}{(x+y)^2}}}$

$grad z=frac y{(x+y)^2sqrt{frac{y(2x+y)}{(x+y)^2}}}bar{i}-frac x{(x+y)^2sqrt{frac{y(2x+y)}{(x+y)^2}}}bar{j}$
$grad z_A=frac 5{(5+5)^2sqrt{frac{5(2cdot5+5)}{(5+5)^2}}}bar{i}-frac 5{(5+5)^2sqrt{frac{5(2cdot5+5)}{(5+5)^2}}}bar{j}=frac5{100sqrt{frac{75}{100}}}bar{i}-frac5{100sqrt{frac{75}{100}}}bar{j}==frac1{10sqrt3}bar{i}-frac1{10sqrt3}bar{j}=frac{sqrt3}{30}bar{i}-frac{sqrt3}{30}bar{j}$

$|grad(z)_A|=sqrt{left(frac{partial z}{partial x}right)^2+left(frac{partial z}{partial y}right)^2}=sqrt{left(frac{sqrt3}{30}right)^2+left(-frac{sqrt3}{30}^2right)^2}=frac{sqrt6}{30}cosalpha=frac{frac{partial z}{partial x}}{|grad{z}|}=frac{frac{sqrt3}{30}}{frac{sqrt6}{30}}=frac1{sqrt2}$ ,
$cosbeta=frac{frac{partial z}{partial y}}{|grad(z)|}=frac{-frac{sqrt3}{30}}{frac{sqrt6}{30}}=-frac1{sqrt2}frac{partial z}{partial a}=frac{partial z}{partial x}cdotcosalpha+frac{partial z}{partial y}cosbeta|a|=sqrt{x^2+y^2}=sqrt{144+25}=13cosalpha=frac x{|a|}=-frac{12}{13}cosbeta=frac y{|a|}=frac5{13}frac{partial z}{partial a}=frac{sqrt3}{30}cdotleft(-frac{12}{13}right)+left(-frac{sqrt3}{30}right)cdotfrac5{13}=-frac{17sqrt3}{390}$