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Question Number 45464 by peter frank last updated on 13/Oct/18

Differentiate with respect to x arctan(((a^2 +x^2 )/(a^2 −x^2 )))

$$\boldsymbol{\mathrm{D}}\mathrm{ifferentiate}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{arctan}}\left(\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right) \\ $$

Commented by peter frank last updated on 14/Oct/18

thank sir.really appreciate

$$\mathrm{thank}\:\mathrm{sir}.\mathrm{really}\:\mathrm{appreciate} \\ $$

Commented by maxmathsup by imad last updated on 14/Oct/18

you are welcome sir

$${you}\:{are}\:{welcome}\:{sir} \\ $$

Commented by maxmathsup by imad last updated on 14/Oct/18

let f(x)=arctan(((a^2 +x^2 )/(a^2 −x^2 ))) =arctan(u(x)) ⇒f^′ (x)=((u^′ (x))/(1+u^2 (x))) but u^′ (x)=((2x(a^2 −x^2 )−(a^2 +x^2 )(−2x))/((a^2 −x^2 )^2 )) =((2a^2 x−2x^3 +2a^2 x +2x^3 )/((a^2 −x^2 )^2 )) =((4a^2 x)/((a^2 −x^2 )^2 )) ⇒f^′ (x) =((4a^2 x)/((a^2 −x^2 )^2 )) .(1/(1+(((a^2 +x^2 )^2 )/((a^2 −x^2 )^2 )))) =((4a^2 x)/((a^2 −x^2 )^2 )) (1/((a^2 −x^2 )^2 +(a^2 +x^2 ))) (a^2 −x^2 )^2 = ((4a^2 x)/(a^4 −2a^2 x^2 +x^4 +a^4 +2a^2 x^(2 ) +x^4 )) =((4a^2 x)/(2a^4 +2x^4 )) ⇒ ★ f^′ (x) =((2a^2 x)/(x^4 +a^4 )) ★ .

$${let}\:{f}\left({x}\right)={arctan}\left(\frac{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)\:={arctan}\left({u}\left({x}\right)\right)\:\Rightarrow{f}^{'} \left({x}\right)=\frac{{u}^{'} \left({x}\right)}{\mathrm{1}+{u}^{\mathrm{2}} \left({x}\right)}\:{but} \\ $$$${u}^{'} \left({x}\right)=\frac{\mathrm{2}{x}\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\left(−\mathrm{2}{x}\right)}{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{2}{a}^{\mathrm{2}} {x}−\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{2}{a}^{\mathrm{2}} {x}\:+\mathrm{2}{x}^{\mathrm{3}} }{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} {x}}{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{\mathrm{4}{a}^{\mathrm{2}} {x}}{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:.\frac{\mathrm{1}}{\mathrm{1}+\frac{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} {x}}{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)}\:\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} {x}}{{a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \:+\mathrm{2}{a}^{\mathrm{2}} {x}^{\mathrm{2}\:} +{x}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} {x}}{\mathrm{2}{a}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{4}} }\:\Rightarrow\:\bigstar\:{f}^{'} \left({x}\right)\:=\frac{\mathrm{2}{a}^{\mathrm{2}} {x}}{{x}^{\mathrm{4}} \:+{a}^{\mathrm{4}} }\:\bigstar\:. \\ $$

Answered by ajfour last updated on 13/Oct/18

let y=tan^(−1) (((a^2 +x^2 )/(a^2 −x^2 ))) (dy/dx)= (1/(1+(((a^2 +x^2 )/(a^2 −x^2 )))^2 ))×((2x(a^2 −x^2 )+2x(a^2 +x^2 ))/((a^2 −x^2 )^2 )) (dy/dx) = ((2a^2 x)/(a^4 +x^4 )) .

$${let}\:\:\:{y}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right) \\ $$$$\frac{{dy}}{{dx}}=\:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }×\frac{\mathrm{2}{x}\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+\mathrm{2}{x}\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{a}^{\mathrm{2}} {x}}{{a}^{\mathrm{4}} +{x}^{\mathrm{4}} }\:\:. \\ $$

Commented by peter frank last updated on 13/Oct/18

thank you sir so much.but why you differentiate numerator only.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\:\mathrm{so}\:\mathrm{much}.\mathrm{but}\:\mathrm{why}\:\mathrm{you}\:\mathrm{differentiate}\:\mathrm{numerator}\:\mathrm{only}. \\ $$

Commented by ajfour last updated on 13/Oct/18

it isn′t so.

$${it}\:{isn}'{t}\:{so}. \\ $$

Commented by peter frank last updated on 14/Oct/18

okay sir i understand now.quetient rule method used.

$$\mathrm{okay}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{now}.\mathrm{quetient}\:\mathrm{rule}\:\mathrm{method}\:\mathrm{used}. \\ $$

Commented by peter frank last updated on 14/Oct/18

sir help Qn 45514

$$\mathrm{sir}\:\mathrm{help}\:\mathrm{Qn}\:\mathrm{45514} \\ $$

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