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Shirley27
 one year ago
How would you take the second derivative of x^2 + 4y^2 =1 for x?
Shirley27
 one year ago
How would you take the second derivative of x^2 + 4y^2 =1 for x?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0simplify it to be in the form: y = f(x) , then differentiate it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Fairly sure now that this is wrongif anyone would like to check my work, feel free and thank you.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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''
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