Question Number 103894 by Study last updated on 18/Jul/20
$$\frac{{d}}{{d}\left(\frac{{d}}{{dx}}{sinx}\right)}\centerdot{sinx}=? \\ $$
Answered by 1549442205 last updated on 18/Jul/20
$$\mathrm{Denote}\:\mathrm{I}=\frac{\mathrm{d}}{\mathrm{d}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}}\mathrm{sinx}=\frac{\mathrm{d}}{\mathrm{dcosx}}\mathrm{sinx}. \\ $$$$\mathrm{Putting}\:\:\mathrm{cos}\:\mathrm{x}=\mathrm{t}.\mathrm{Then} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{I}=\frac{\mathrm{d}}{\mathrm{dt}}\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }=\frac{−\mathrm{t}}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}=\frac{−\mathrm{cosx}}{\mathrm{sinx}} \\ $$$$=−\mathrm{cot}\:\mathrm{x} \\ $$$$ \\ $$
Answered by MAB last updated on 18/Jul/20
$$ \\ $$$$\frac{{d}}{{d}\left(\frac{{d}}{{dx}}{sinx}\right)}\centerdot{sinx}=\frac{{d}}{{dcos}\left({x}\right)}{sin}\left({x}\right) \\ $$$$=\frac{{dsin}\left({x}\right)}{{dx}}\centerdot\frac{{dx}}{{dcos}\left({x}\right)} \\ $$$$={cos}\left({x}\right)\centerdot\left(−\frac{\mathrm{1}}{{sin}\left({x}\right)}\right) \\ $$$$=−{cot}\left({x}\right) \\ $$$$ \\ $$
Commented by Study last updated on 18/Jul/20
$${how}\:{we}\:{can}\:{calculate}\:\frac{{dx}}{{dcos}\left({x}\right)}? \\ $$
Commented by bobhans last updated on 18/Jul/20
$$\frac{{dx}}{{d}\left(\mathrm{cos}\:{x}\right)}\:=\:\frac{\mathrm{1}}{\frac{{d}\left(\mathrm{cos}\:{x}\right)}{{dx}}}\:=\:\frac{\mathrm{1}}{−\mathrm{sin}\:{x}} \\ $$
Answered by bobhans last updated on 18/Jul/20
$$\frac{{d}\left(\mathrm{sin}\:{x}\right)}{{d}\left(\frac{{d}\left(\mathrm{sin}\:{x}\right)}{{dx}}\right)}\:=\:\frac{{d}\left(\mathrm{sin}\:{x}\right)}{{d}\left(\mathrm{cos}\:{x}\right)}\:=\:\frac{{d}\left(\mathrm{sin}\:{x}\right)}{{dx}}\:×\:\frac{{dx}}{{d}\left(\mathrm{cos}\:{x}\right)} \\ $$$$=\:\mathrm{cos}\:{x}\:×\:\frac{\mathrm{1}}{\frac{{d}\left(\mathrm{cos}\:{x}\right)}{{dx}}}\:=\:\mathrm{cos}\:{x}\:×\frac{\mathrm{1}}{−\mathrm{sin}\:{x}} \\ $$$$=\:−\mathrm{cot}\:{x}\: \\ $$