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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
ify=sin1x2+cos1x2finddy/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(x)=arcsin(x2)+arcos(x2)y(x)=2x1x4+2x1x4=0y(x)=λxRy(0)=π2y(x)=π2.
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  .
ifyoumeany(x)=(arcsinx)2+(arcosx)2y(x)=2arcsinx1x22arcosx1x2=21x2(arcsinxarcosx)=21x2(arcos(x)+arcsin(x))=21x2ξπ2=πξ1x2withξ2=1and1<x<1.
Commented by NECx last updated on 24/Jan/18
thank you so much.
thankyousomuch.

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