Shirley27
  • Shirley27
How would you take the second derivative of x^2 + 4y^2 =1 for x?
OCW Scholar - Single Variable Calculus
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Shirley27
  • Shirley27
How would you take the second derivative of x^2 + 4y^2 =1 for x?
OCW Scholar - Single Variable Calculus
schrodinger
  • schrodinger
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anonymous
  • anonymous
simplify it to be in the form: y = f(x) , then differentiate it
anonymous
  • anonymous
You could differentiate the whole equation: f'(x^2 + 4y^2 = 1) = (2x + 8y*y' = 0) 8y*y' = -2x => y'=-2x/8y and then differentiate again: y''=f'(-2x/8y) (using quotient rule): (-2*1*8y-(-2x*8y'))/(64y^2) = y'' (-16y+16xy')/(64y^2) = y'' (-4y+4x*(-2x/8y))/y^2 = y'' (-4y/y^2)+(-8x/8y^3) = y'' -4/y + x/y^3 = y'' whew. that feels wrong. This will likely be a learning experience for me as well! Thanks.
Shirley27
  • Shirley27
Thanks!!

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anonymous
  • anonymous
Fairly sure now that this is wrong--if anyone would like to check my work, feel free and thank you.
anonymous
  • anonymous
the approach shulmand used is correct, but missed some accuracy. Here's the correct version. (using quotient rule): (-2*1*8y-(-2x*8y'))/(64y^2) = y'' (-16y+16xy')/(64y^2) = y'' (-4y+4x*(-2x/8y))/(16y^2) = y'' (-4y/16y^2)+(-8x^2/(16*8y^3)) = y'' -1/4y - x^2/16y^3 = y''
anonymous
  • anonymous
Thank you.

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