Question Number 124845 by liberty last updated on 06/Dec/20
$$\:\:\:{f}\left({x}\right)\:=\:\mathrm{sin}\:\left(\frac{{x}}{{x}−\mathrm{sin}\:\left(\frac{{x}}{{x}−\mathrm{sin}\:{x}}\right)}\right)\: \\ $$$$\:\:\frac{{df}\left({x}\right)}{{dx}}\:? \\ $$
Answered by bemath last updated on 06/Dec/20
$${f}\:'\left({x}\right)=\frac{\mathrm{1}\left({x}−\mathrm{sin}\:\left(\frac{{x}}{{x}−\mathrm{sin}\:{x}}\right)\right)−{x}\left(\mathrm{1}−\left(\frac{{x}−\mathrm{sin}\:{x}−{x}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{\left({x}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\right)\mathrm{cos}\:\left(\frac{{x}}{{x}−\mathrm{sin}\:{x}}\right)\right.}{\left({x}−\mathrm{sin}\:\left(\frac{{x}}{{x}−\mathrm{sin}\:{x}}\right)\right)^{\mathrm{2}} }.\:\mathrm{cos}\:\left(\frac{{x}}{{x}−\mathrm{sin}\:\left(\frac{{x}}{{x}−\mathrm{sin}\:{x}}\right)}\right) \\ $$
Answered by mindispower last updated on 06/Dec/20
$${g}\left({x}\right)=\frac{{x}}{{x}−{sin}\left({x}\right)} \\ $$$${arcsin}\left({fx}\right)=\frac{{x}}{{x}−{sin}\left({g}\left({x}\right)\right)} \\ $$$$\Rightarrow\frac{{f}'}{\:\sqrt{\mathrm{1}−{f}^{\mathrm{2}} }}=\frac{−{sin}\left({g}\left({x}\right)\right)+{g}'{cos}\left({g}\left({x}\right)\right)}{{x}−{sin}\left({g}\left({x}\right)\right)^{\mathrm{2}} }\:_{={A}} \\ $$$${just}\:{sam}\:{as}\:{A}\:{withe}\:{g}={x},{g}'=\mathrm{1} \\ $$$${g}'=\frac{−{sin}\left({x}\right)+{cos}\left({x}\right)}{\left({x}−{sin}\left({x}\right)\right)^{\mathrm{2}} } \\ $$