x^3=y^2; 2nd derivative, quotient rule

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colemma

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I am in 12th grade, have a question about a 2nd derivative that requires use of the quotient rule. I thought I could do it, but the published answer is different.

Find 2nd derivative of x^3=y^2

x^3-y^2=0

derivative

3x^2 - 2y dy/dx = 0

dy/dx = 3x^2/2y

then using quotient rule, d2y/dx2 is derivative of 3x^2/2y

(6x.2y- 3x^2.2dy/dx)/(2y)^2

(12xy - 6x^2.3x^2/2y)/(2y)^2

12xy/4y^2 - 18x^4/2y/4y^2

3x/y - 9x^4/4y^3

Which seems OK

But answer sheet says 12xy - 9x^4/4y^3

So clearly I lost (2y)^2 from first term, but I don't know how.
 
Except for missing parentheses, the answer sheet's (12xy - 9x^4)/4y^3 is exactly what you have before you broke into two fractions.
 
colemma said:
...

x^3-y^2=0
...
y'' = 3x/y - 9x^4/4y^3

Which seems OK
...
Click to expand...

Sorry, got interupted. To continue, let's go about this a different way:

From the original
x^3-y^2=0
we have
y = x3/2
y' = \(\displaystyle \frac{3}{2}x^{1/2}\)
y'' = \(\displaystyle \frac{3}{4}x^{-1/2}\)

y'' = \(\displaystyle \frac{3x}{y} - \frac{9x^4}{4y^3} = \frac{3x}{x^{3/2}} - \frac{9x^4}{4x^{9/2}}\)
= \(\displaystyle \frac{12}{4x^{1/2}} - \frac{9}{4x^{1/2}} \) = \(\displaystyle \frac{3}{4}x^{-1/2}\)

So looks to me like your answer is correct.
 
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