Question Number 226879 by Spillover last updated on 17/Dec/25
Answered by gregori last updated on 18/Dec/25
$$\left({a}\right)\:{f}\:'\left({x}\right)=\:−\mathrm{sin}\:{x}\:.\:{e}^{\mathrm{cos}\:{x}} \: \\ $$$$\:\left({b}\right)\:{f}\:'\left({x}\right)\:=\:\mathrm{3sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:{x}\:.{e}^{\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$ \\ $$
Commented by Spillover last updated on 18/Dec/25
$${thanks} \\ $$
Answered by mr W last updated on 18/Dec/25
$$\left({a}\right)\:{f}'\left({x}\right)=−\mathrm{sin}\:{x}\:{e}^{\mathrm{cos}\:{x}} \\ $$$$ \\ $$$$\left({b}\right)\: \\ $$$$\mathrm{ln}\:{f}\left({x}\right)=\mathrm{sin}^{\mathrm{3}} \:{x}\:\mathrm{ln}\:\mathrm{2} \\ $$$$\frac{{f}'\left({x}\right)}{{f}\left({x}\right)}=\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}\:{x}\:\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow{f}'\left({x}\right)=\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}\:{x}\:\mathrm{ln}\:\mathrm{2}\:\mathrm{2}^{\mathrm{sin}^{\mathrm{3}} \:{x}} \\ $$
Commented by Spillover last updated on 18/Dec/25
$${thanks} \\ $$