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Inverse trig

  • Thread starter nameVoid
  • Start date
x^2+xsin^-1y=ye^x

implict dif

2x+xy'/(1-x^2)^(1/2)+sin^(-1)y=ye^x+y'e^x
xy'/(1-x^2)^(1/2)-y'e^x=ye^x-sin^(-1)y-2x
y'=(ye^x-sin^(-1)y-2x)/(x/(1-x^2)^(1/2)-e^x)
can somone please verify
 

ideasrule

Homework Helper
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x^2+xsin^-1y=ye^x
2x+xy'/(1-x^2)^(1/2)+sin^(-1)y=ye^x+y'e^x
You're differentiating sin^-1y with respect to x, so you use the chain rule and first differentiate with respect to y, then derive y with respect to x, then multiply the two derivatives together. That would give you 1/sqrt(1-y^2) * dy/dx, not 1/sqrt(1-x^2) *dy/dx
 
ah yes that was a typeo on my part my solution is

(ye^x-sin^-1y-2x)/(x/(1-y^2)^(1/2)-e^x)
my book is showing
(ye^x-2x-sin^-1y)/(x/((1-y^2)^(1/2)-e^x))

which divides (1-y^2)^(1/2)-e^x into x not just x/(1-y^2)^(1/2) i dont believe these solutions are equal
 

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