Lexi Mcneil

2022-07-23

Find the derivative $\frac{d}{dx}\left[\mathrm{arctan}\left({x}^{2}+1\right)\right]$
$\left(arcsec\left({x}^{2}+1\right){\right)}^{2}\phantom{\rule{0ex}{0ex}}\frac{1}{2+{x}^{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{1+\left({x}^{2}+{\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{2x}{1+\left({x}^{2}+1{\right)}^{2}}$

Anaya Gregory

Expert

Find the derivative of $\left[\mathrm{arctan}\left({x}^{2}+1\right)\right]$
$f\left(x\right)=\mathrm{arctan}\left(1+{x}^{2}\right)⇒{\mathrm{tan}}^{-1}\left(1+{x}^{2}\right)\phantom{\rule{0ex}{0ex}}f\left(x\right)={\mathrm{tan}}^{-1}\left(1+{x}^{2}\right)\left[\text{diffrenciate both side}\right]\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left({\mathrm{tan}}^{-1}\left(1+{x}^{2}\right)\right)\left\{\begin{array}{r}\frac{d}{dx}\left({\mathrm{tan}}^{-1}\left(x\right)\right)⇒\frac{1}{1+{x}^{2}}\\ \frac{d}{dx}\left({x}^{n}\right)⇒n{x}^{\left(n-1\right)}\end{array}\right\}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{1}{\left[1+\left(1+{x}^{2}{\right)}^{2}\right]}\frac{d}{dx}\left(1+{x}^{2}\right)\left\{\frac{d}{dx}\left(constant\right)=0\right\}\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)⇒\frac{1\left(0+2x\right)}{1+\left(1+{x}^{2}{\right)}^{2}}⇒\frac{2x}{1+\left(1+{x}^{2}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}Hence,\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left[\mathrm{arctan}\left(1+{x}^{2}\right)\right]⇒\frac{2x}{1+\left({x}^{2}+1{\right)}^{2}}$

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