if-y-sin-1-x-2-cos-1-x-2-find-dy-dx-

Question Number 28304 by NECx last updated on 23/Jan/18

$${if}\:{y}=\left(\mathrm{sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} +\left(\mathrm{cos}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${find}\:{dy}/{dx} \\ $$

Answered by ajfour last updated on 23/Jan/18

$${let}\:\:\:\mathrm{sin}^{−\mathrm{1}} {x}=\theta\:\:,\:\Rightarrow\:\frac{{d}\theta}{{dx}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${y}=\theta^{\:\mathrm{2}} +\left(\frac{\pi}{\mathrm{2}}−\theta\right)^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\left[\mathrm{2}\theta−\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\theta\right)\right]\frac{{d}\theta}{{dx}} \\ $$$$\:\frac{{dy}}{{dx}}\:=−\frac{\pi}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:. \\ $$

Commented by NECx last updated on 24/Jan/18

$${I}\:{took}\:{the}\:{following}\:{steps}\:{but}\:{I} \\ $$$${dont}\:{know}\:{if}\:{my}\:{analysis}\:{is}\:{wrong} \\ $$$${please}\:{check}\:{and}\:{make}\:{the} \\ $$$${possible}\:{corrections} \\ $$$$ \\ $$$${y}=\left(\mathrm{sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} +\left(\mathrm{cos}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} \\ $$$${let}\:\mathrm{sin}^{−\mathrm{1}} {x}={a}=\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$\therefore{y}=\left(\mathrm{cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\right)^{\mathrm{2}} +\left(\mathrm{cos}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{−\mathrm{2cos}^{−\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+\:\frac{\left(−\mathrm{2cos}^{−\mathrm{1}} {x}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}=\frac{−\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\mathrm{cos}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}+\:\mathrm{cos}^{−\mathrm{1}} {x}\right) \\ $$$$ \\ $$$${dy}/{dx}=\frac{−\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\mathrm{0}\right) \\ $$$$ \\ $$$${then}\:{dy}/{dx}=\mathrm{0} \\ $$

Commented by mrW2 last updated on 24/Jan/18

$$\mathrm{cos}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}+\:\mathrm{cos}^{−\mathrm{1}} {x} \\ $$$$=\mathrm{sin}^{−\mathrm{1}} {x}+\:\mathrm{cos}^{−\mathrm{1}} {x} \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$$$\neq\mathrm{0} \\ $$

Commented by NECx last updated on 24/Jan/18

$${oh}\:{Thanks} \\ $$

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