Question Number 93725 by ckkim89 last updated on 14/May/20
Answered by Rio Michael last updated on 14/May/20
$$\mathrm{let}\:{y}\:=\:{f}\left({x}\right)\:\Rightarrow\:{y}\:=\:\mathrm{cos}^{−\mathrm{1}} \left(−\mathrm{3}{x}\right) \\ $$$$\Rightarrow\:\mathrm{cos}\:{y}\:=\:−\mathrm{3}{x} \\ $$$$\Rightarrow\:−\mathrm{sin}\:{y}\:{y}\:'\:=\:−\mathrm{3} \\ $$$$\Rightarrow\:{y}'\:=\:\frac{\mathrm{3}}{\mathrm{sin}\:{y}} \\ $$$$\mathrm{but}\:\mathrm{sin}\:{y}\:=\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {y}} \\ $$$$\:\Rightarrow\:{y}'\:=\:\frac{\mathrm{3}}{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:{f}\:'\left({x}\right)\:=\:\frac{\mathrm{3}}{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }} \\ $$
Commented by ckkim89 last updated on 14/May/20
thanks!
Answered by hknkrc46 last updated on 15/May/20
![f(x)=y=cos^(−1) (−3x)⇒y^′ =−(((−3x)^′ )/(√(1−(−3x)^2 ))) =(3/(√(1−9x^2 ))) (d/dx)cos^(−1) [h(x)]=−((h^′ (x))/(√(1−h^2 (x))))](Q93800.png)
$${f}\left({x}\right)={y}=\mathrm{cos}^{−\mathrm{1}} \left(−\mathrm{3}{x}\right)\Rightarrow{y}^{'} =−\frac{\left(−\mathrm{3}{x}\right)^{'} }{\sqrt{\mathrm{1}−\left(−\mathrm{3}{x}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{3}}{\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} }} \\ $$$$\frac{{d}}{{dx}}\mathrm{cos}^{−\mathrm{1}} \left[{h}\left({x}\right)\right]=−\frac{{h}^{'} \left({x}\right)}{\sqrt{\mathrm{1}−{h}^{\mathrm{2}} \left({x}\right)}} \\ $$