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Question-48501




Question Number 48501 by peter frank last updated on 24/Nov/18
Answered by Abdulhafeez Abu qatada last updated on 25/Nov/18
  ((df/dx))_y  = sinx(sin(x+y)cosy + cos(x+y)siny)  ((df/dx))_y  = sinx(sin(x+y)cosy + cos(x+y)siny)  ((df/dx))_y  = sinx(sin(x+y)+(y))  ((df/dx))_y  = sinx(sin(x+2y)      ((df/dx))_x  = siny(sin(x+y)cosx + cos(x+y)sinx)  ((df/dx))_x  = siny(sin(x+y + (x)))  ((df/dx))_x  = siny(sin(y+2x)    Abu qatada
$$ \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{y}} \:=\:\mathrm{sin}{x}\left(\mathrm{sin}\left({x}+{y}\right)\mathrm{cos}{y}\:+\:\mathrm{cos}\left({x}+{y}\right)\mathrm{sin}{y}\right) \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{y}} \:=\:\mathrm{sin}{x}\left(\mathrm{sin}\left({x}+{y}\right)\mathrm{cos}{y}\:+\:\mathrm{cos}\left({x}+{y}\right)\mathrm{sin}{y}\right) \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{y}} \:=\:\mathrm{sin}{x}\left(\mathrm{sin}\left({x}+{y}\right)+\left({y}\right)\right) \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{y}} \:=\:\mathrm{sin}{x}\left(\mathrm{sin}\left({x}+\mathrm{2}{y}\right)\right. \\ $$$$ \\ $$$$ \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{x}} \:=\:\mathrm{sin}{y}\left(\mathrm{sin}\left({x}+{y}\right)\mathrm{cos}{x}\:+\:\mathrm{cos}\left({x}+{y}\right)\mathrm{sin}{x}\right) \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{x}} \:=\:\mathrm{sin}{y}\left(\mathrm{sin}\left({x}+{y}\:+\:\left({x}\right)\right)\right) \\ $$$$\left(\frac{{df}}{{dx}}\right)_{{x}} \:=\:\mathrm{sin}{y}\left(\mathrm{sin}\left({y}+\mathrm{2}{x}\right)\right. \\ $$$$ \\ $$$${Abu}\:{qatada} \\ $$

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