Question

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

Answer 1

simplify it to be in the form: y = f(x) , then differentiate it

Answer 2

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.

Answer 3

Thanks!!

Answer 4

Fairly sure now that this is wrong--if anyone would like to check my work, feel free and thank you.

Answer 5

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''

Answer 6

Thank you.