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

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

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

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

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

$$youarewelcomesir$$

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

$$letf\left(x\right)=arctan\left(\frac{a^2+x^2}{a^2-x^2}\right)=arctan\left(u\left(x\right)\right)\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{u^{{}^{\prime }}\left(x\right)}{1+u^2\left(x\right)}but$$$$u^{{}^{\prime }}\left(x\right)=\frac{2x(a^2-x^2)-(a^2+x^2)(-2x)}{{(a^2-x^2)}^2}=\frac{2a^{2}x-2x^3+2a^{2}x+2x^3}{{(a^2-x^2)}^2}$$$$=\frac{4a^{2}x}{{(a^2-x^2)}^2}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{4a^{2}x}{{(a^2-x^2)}^2}.\frac{1}{1+\frac{{(a^2+x^2)}^2}{{(a^2-x^2)}^2}}$$$$=\frac{4a^{2}x}{{(a^2-x^2)}^2}\frac{1}{{(a^2-x^2)}^2+(a^2+x^2)}{(a^2-x^2)}^2=\frac{4a^{2}x}{a^4-2a^2{x}^2+x^4+a^4+2a^2{x}^2+x^4}$$$$=\frac{4a^{2}x}{2a^4+2x^4}\Rightarrow ★f^{{}^{\prime }}\left(x\right)=\frac{2a^{2}x}{x^4+a^4}★.$$

Answered by ajfour last updated on 13/Oct/18

$$lety={\tan }^{-1}\left(\frac{a^2+x^2}{a^2-x^2}\right)$$$$\frac{dy}{dx}=\frac{1}{1+{\left(\frac{a^2+x^2}{a^2-x^2}\right)}^2}\times \frac{2x(a^2-x^2)+2x(a^2+x^2)}{{(a^2-x^2)}^2}$$$$\frac{dy}{dx}=\frac{2a^{2}x}{a^4+x^4}.$$

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

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

Commented by ajfour last updated on 13/Oct/18

$$it{isn}^{\prime }tso.$$

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

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

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

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

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