Question Number 81565 by jagoll last updated on 14/Feb/20
$${If}\:{f}\left({x}\right)=\:\left(\mathrm{tan}\:{x}\right)^{\mathrm{cot}\:{x}} \:+\:\left(\mathrm{cot}\:{x}\right)^{\mathrm{tan}\:{x}} \\ $$$${f}\:'\left({x}\right)=\:? \\ $$
Commented by john santu last updated on 14/Feb/20
$$\mathrm{good}\:\mathrm{question} \\ $$
Commented by mathmax by abdo last updated on 14/Feb/20
$${f}\left({x}\right)=\left({tanx}\right)^{{cotanx}} \:+\left({cotanx}\right)^{{tanx}} \:\Rightarrow \\ $$$${f}\left({x}\right)={e}^{\frac{\mathrm{1}}{{tanx}}{ln}\left({tanx}\right)} \:+{e}^{{tanxln}\left(\frac{\mathrm{1}}{{tanx}}\right)} \:={e}^{\frac{{ln}\left({tanx}\right)}{{tanx}}} \:+{e}^{−{tanxln}\left({tanx}\right)} \\ $$$$={u}\left({x}\right)+{v}\left({x}\right)\:\Rightarrow{f}^{'} \left({x}\right)={u}^{'} \left({x}\right)+{v}^{'} \left({x}\right) \\ $$$${u}^{'} \left({x}\right)=\left(\frac{{ln}\left({tanx}\right)}{{tanx}}\right)^{'} {u}\left({x}\right)\:=\left\{\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)}{{tanx}}×{tanx}\:−{ln}\left({tanx}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\right\}×\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {x}}×{u}\left({x}\right) \\ $$$$=\left\{\mathrm{1}+{tan}^{\mathrm{2}} {x}−{ln}\left({tanx}\right)−{ln}\left({tanx}\right){tan}^{\mathrm{2}} {x}\right\}×\frac{{u}\left({x}\right)}{{tan}^{\mathrm{2}} {x}} \\ $$$${v}^{'} \left({x}\right)=\left(−{tanxln}\left({tanx}\right)\right)^{'} {v}\left({x}\right) \\ $$$$=−\left\{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){ln}\left({tanx}\right)+{tanx}×\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{{tanx}}\right\}{v}\left({x}\right) \\ $$$$=−\left\{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){ln}\left({tanx}\right)+\mathrm{1}+{tan}^{\mathrm{2}} {x}\right\}{v}\left({x}\right) \\ $$$$ \\ $$
Answered by john santu last updated on 14/Feb/20
Commented by jagoll last updated on 14/Feb/20
$${thank}\:{a}\:{lot}\:{of} \\ $$