Question Number 74933 by imelder last updated on 04/Dec/19
$$\mathrm{differentiate}\:\mathrm{the}\:\mathrm{following}\:\mathrm{functions} \\ $$$$\left.\mathrm{a}\right)\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}^{\mathrm{5}} \:\mathrm{coshx} \\ $$
Commented by mathmax by abdo last updated on 04/Dec/19
$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{5}} {ch}\left({x}\right)=\mathrm{2}{x}^{\mathrm{5}} ×\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}\:={x}^{\mathrm{5}} \left({e}^{{x}} \:+{e}^{−{x}} \right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\mathrm{5}{x}^{\mathrm{4}} \left({e}^{{x}} +{e}^{−{x}} \right)\:+{x}^{\mathrm{5}} \left({e}^{{x}} −{e}^{−{x}} \right)\:=\left(\mathrm{5}{x}^{\mathrm{4}} \:+{x}^{\mathrm{5}} \right){e}^{{x}} \:+\left(\mathrm{5}{x}^{\mathrm{4}} −{x}^{\mathrm{5}} \right){e}^{−{x}} \\ $$$${another}\:{way}\:\:{f}^{'} \left({x}\right)=\mathrm{10}{x}^{\mathrm{4}} {ch}\left({x}\right)+\mathrm{2}{x}^{\mathrm{5}} {sh}\left({x}\right) \\ $$