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Question Number 81565 by jagoll last updated on 14/Feb/20

If f(x)= (tan x)^(cot x) + (cot x)^(tan x) f ′(x)= ?

$${If}\:{f}\left({x}\right)=\:\left(\mathrm{tan}\:{x}\right)^{\mathrm{cot}\:{x}} \:+\:\left(\mathrm{cot}\:{x}\right)^{\mathrm{tan}\:{x}} \\ $$$${f}\:'\left({x}\right)=\:? \\ $$

Commented by john santu last updated on 14/Feb/20

good question

$$\mathrm{good}\:\mathrm{question} \\ $$

Commented by mathmax by abdo last updated on 14/Feb/20

f(x)=(tanx)^(cotanx) +(cotanx)^(tanx) ⇒ f(x)=e^((1/(tanx))ln(tanx)) +e^(tanxln((1/(tanx)))) =e^((ln(tanx))/(tanx)) +e^(−tanxln(tanx)) =u(x)+v(x) ⇒f^′ (x)=u^′ (x)+v^′ (x) u^′ (x)=(((ln(tanx))/(tanx)))^′ u(x) ={(((1+tan^2 x))/(tanx))×tanx −ln(tanx)(1+tan^2 x)}×(1/(tan^2 x))×u(x) ={1+tan^2 x−ln(tanx)−ln(tanx)tan^2 x}×((u(x))/(tan^2 x)) v^′ (x)=(−tanxln(tanx))^′ v(x) =−{(1+tan^2 x)ln(tanx)+tanx×((1+tan^2 x)/(tanx))}v(x) =−{(1+tan^2 x)ln(tanx)+1+tan^2 x}v(x)

$${f}\left({x}\right)=\left({tanx}\right)^{{cotanx}} \:+\left({cotanx}\right)^{{tanx}} \:\Rightarrow \\ $$$${f}\left({x}\right)={e}^{\frac{\mathrm{1}}{{tanx}}{ln}\left({tanx}\right)} \:+{e}^{{tanxln}\left(\frac{\mathrm{1}}{{tanx}}\right)} \:={e}^{\frac{{ln}\left({tanx}\right)}{{tanx}}} \:+{e}^{−{tanxln}\left({tanx}\right)} \\ $$$$={u}\left({x}\right)+{v}\left({x}\right)\:\Rightarrow{f}^{'} \left({x}\right)={u}^{'} \left({x}\right)+{v}^{'} \left({x}\right) \\ $$$${u}^{'} \left({x}\right)=\left(\frac{{ln}\left({tanx}\right)}{{tanx}}\right)^{'} {u}\left({x}\right)\:=\left\{\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)}{{tanx}}×{tanx}\:−{ln}\left({tanx}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\right\}×\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {x}}×{u}\left({x}\right) \\ $$$$=\left\{\mathrm{1}+{tan}^{\mathrm{2}} {x}−{ln}\left({tanx}\right)−{ln}\left({tanx}\right){tan}^{\mathrm{2}} {x}\right\}×\frac{{u}\left({x}\right)}{{tan}^{\mathrm{2}} {x}} \\ $$$${v}^{'} \left({x}\right)=\left(−{tanxln}\left({tanx}\right)\right)^{'} {v}\left({x}\right) \\ $$$$=−\left\{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){ln}\left({tanx}\right)+{tanx}×\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{{tanx}}\right\}{v}\left({x}\right) \\ $$$$=−\left\{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){ln}\left({tanx}\right)+\mathrm{1}+{tan}^{\mathrm{2}} {x}\right\}{v}\left({x}\right) \\ $$$$ \\ $$

Answered by john santu last updated on 14/Feb/20

Commented by jagoll last updated on 14/Feb/20

thank a lot of

$${thank}\:{a}\:{lot}\:{of} \\ $$

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