Lexi Mcneil

Answered

2022-07-23

Find the derivative $\frac{d}{dx}[\mathrm{arctan}({x}^{2}+1)]$

$(arcsec({x}^{2}+1){)}^{2}\phantom{\rule{0ex}{0ex}}\frac{1}{2+{x}^{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{1+({x}^{2}+{)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{2x}{1+({x}^{2}+1{)}^{2}}$

$(arcsec({x}^{2}+1){)}^{2}\phantom{\rule{0ex}{0ex}}\frac{1}{2+{x}^{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{1+({x}^{2}+{)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{2x}{1+({x}^{2}+1{)}^{2}}$

Answer & Explanation

Anaya Gregory

Expert

2022-07-24Added 14 answers

Find the derivative of $[\mathrm{arctan}({x}^{2}+1)]$

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

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

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