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