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Question Number 145021 by mim24 last updated on 01/Jul/21

Answered by phally last updated on 01/Jul/21

$$\mathrm{Derivative}\mathrm{formula}$$$${\left({\mathrm{Cos}}^{-1}\left(\mathrm{u}\right)\right)}^{{}^{\prime }}=\frac{-{\mathrm{u}}^{\prime }}{\sqrt{1-{\mathrm{u}}^2}}$$

Answered by mathmax by abdo last updated on 01/Jul/21

$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{arcosx}\left(\frac{\mathrm{x}-{\mathrm{x}}^{-1}}{\mathrm{x}+{\mathrm{x}}^{-1}}\right)=\mathrm{arcos}\left(\frac{\mathrm{x}-\frac{1}{\mathrm{x}}}{\mathrm{x}+\frac{1}{\mathrm{x}}}\right)=\mathrm{arcos}\left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\right)\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{{\left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\right)}^{{}^{\prime }}}{\sqrt{1-{\left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\right)}^2}}\mathrm{but}{\left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\right)}^{{}^{\prime }}={(1-\frac{2}{{\mathrm{x}}^2+1})}^{{}^{\prime }}=2\times \frac{2\mathrm{x}}{{({\mathrm{x}}^2+1)}^2}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4\mathrm{x}}{{({\mathrm{x}}^2+1)}^2\sqrt{1-\frac{{({\mathrm{x}}^2-1)}^2}{{({\mathrm{x}}^2+1)}^2}}}=\frac{4\mathrm{x}}{{({\mathrm{x}}^2+1)}^2\frac{\sqrt{{({\mathrm{x}}^2+1)}^2-{({\mathrm{x}}^2-1)}^2}}{{\mathrm{x}}^2+1}}$$$$=\frac{4\mathrm{x}}{({\mathrm{x}}^2+1)\sqrt{{\mathrm{x}}^4+{2\mathrm{x}}^2+1-{\mathrm{x}}^4+{2\mathrm{x}}^2-1}}=\frac{4\mathrm{x}}{({\mathrm{x}}^2+1)\sqrt{{4\mathrm{x}}^2}}$$$$=\frac{4\mathrm{x}}{2\mid \mathrm{x}\mid ({\mathrm{x}}^2+1)}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\xi \left(\mathrm{x}\right)}{{\mathrm{x}}^2+1}\mathrm{with}\xi \left(\mathrm{x}\right)=1\mathrm{if}\mathrm{x}>0\mathrm{and}\xi \left(\mathrm{x}\right)=-1\mathrm{if}\mathrm{x}<0$$

Commented by mathmax by abdo last updated on 01/Jul/21

$$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2\xi \left(\mathrm{x}\right)}{{\mathrm{x}}^2+1}$$

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