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Question Number 28305 by NECx last updated on 23/Jan/18
if y=sin^(−1) x^2 +cos^(−1) x^2   find dy/dx
$${if}\:{y}=\mathrm{sin}^{−\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{cos}^{−\mathrm{1}} {x}^{\mathrm{2}} \\ $$$${find}\:{dy}/{dx} \\ $$
Commented by abdo imad last updated on 23/Jan/18
y(x)= arcsin(x^2 ) +arcos(x^2 )  y^′ (x)=  ((2x)/( (√(1−x^4 ))))  +((−2x)/( (√(1−x^4 )))) =0 ⇒y(x)= λ  ∀ x ∈ R  y(0)= (π/2)⇒  y(x)= (π/2)  .
$${y}\left({x}\right)=\:{arcsin}\left({x}^{\mathrm{2}} \right)\:+{arcos}\left({x}^{\mathrm{2}} \right) \\ $$$${y}^{'} \left({x}\right)=\:\:\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\:+\frac{−\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:=\mathrm{0}\:\Rightarrow{y}\left({x}\right)=\:\lambda\:\:\forall\:{x}\:\in\:{R} \\ $$$${y}\left(\mathrm{0}\right)=\:\frac{\pi}{\mathrm{2}}\Rightarrow\:\:{y}\left({x}\right)=\:\frac{\pi}{\mathrm{2}}\:\:. \\ $$
Commented by abdo imad last updated on 23/Jan/18
if you mean y(x)= (arcsinx)^2 + (arcosx)^2   y^′ (x)=2((arcsinx)/( (√(1−x^2 )))) −2((arcosx)/( (√(1−x^2 ))))=(2/( (√(1−x^2 ))))( arcsinx −arcosx)  = ((−2)/( (√(1−x^2 ))))(arcos(−x) +arcsin(−x))=((−2)/( (√(1−x^2 )))) ((ξπ)/2)  =((−πξ)/( (√(1−x^2 ))))  with  ξ^2 =1   and −1<x<1  .
$${if}\:{you}\:{mean}\:{y}\left({x}\right)=\:\left({arcsinx}\right)^{\mathrm{2}} +\:\left({arcosx}\right)^{\mathrm{2}} \\ $$$${y}^{'} \left({x}\right)=\mathrm{2}\frac{{arcsinx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{2}\frac{{arcosx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\:{arcsinx}\:−{arcosx}\right) \\ $$$$=\:\frac{−\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left({arcos}\left(−{x}\right)\:+{arcsin}\left(−{x}\right)\right)=\frac{−\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\frac{\xi\pi}{\mathrm{2}} \\ $$$$=\frac{−\pi\xi}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{with}\:\:\xi^{\mathrm{2}} =\mathrm{1}\:\:\:{and}\:−\mathrm{1}<{x}<\mathrm{1}\:\:. \\ $$
Commented by NECx last updated on 24/Jan/18
thank you so much.
$${thank}\:{you}\:{so}\:{much}. \\ $$

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