Second Derivative (Implicit Differentiation)

[quicksilver123]

Homework Statement

Find y''

Homework Equations

9x^2 +y^2 = 9

The Attempt at a Solution

y'
18x+2y(y')=0
y'=-18x/2y
y'=9x/y

For the second derivative, I get the correct answer (same as the book) up until the very last step.
Here's where I'm left at:

-9( (-9x^2 - y^2) / y^3 )

The book then takes this answer to:

y''= -9 (9/y^3) [since x and y must satisfy the original equation, 9x^2+y^2=9] Thus y''= - 81/y^3

I'm not sure how they got here as my final answer ends up something like -81x^2/y^3 -9y (probably with some sign mistakes in there but I was too distracted with the book's justification for their answer to worry about that).

Help?

 

[fresh_42]

I think you have a sign error somewhere. As you haven't shown your intermediate steps, I cannot say where. Only that your last equation for $y'$ is wrong. Now, why are you stuck? What is $(-9x^2-y^2)$?

 

[quicksilver123]

Alright I'll show my intermediate steps for the first to second derivative.

Y'=-9x/y

Y''=(-9y-(-9x)y')/y^2
Y''=-9(y+9xy')/y^2
Y''=-9(y+9x(-9x/y))/y^2
Y''=-9(y/y)(y+9x(-9x/y))/y^2
Y''=-9(y^2-81x^2)/y^3
Y''=-9y^2/y^3 -9(-81x^2)/y^3
Y''=-9/y + 9(81x^2)/y^3

Slightly different from my end answer but still no closer to the book's.
Apologies for the plaintext as I am on mobile.

 

[fresh_42]

I don't understand your third row. Did you pull $-9$ out of the parentheses or not? And there is still a sign error.

 

[quicksilver123]

[QUOTE="quicksilver123, post: 5594654, member
Y'=-9x/y

Y''=(-9y-(-9x)y')/y^2
Y''=-9(y+xy')/y^2
Y''=-9(y+x(-9x/y))/y^2
Y''=-9(y/y)(y+x(-9x/y))/y^2
Y''=-9(y^2-9x^2)/y^3
Y''=-9y^2/y^3 -9(-9x^2)/y^3
Y''=-9/y + 81x^2/y^3

Slightly different from my end answer but still no closer to the book's.Made some corrections.

 

[fresh_42]

This sign error is pretty resistant. May have to do with your linear notation.
You have $y''=(-9y-(-9x)y')/y^2$ which is
$$y''=\frac{-9y-(-9x)y'}{y^2}=\frac{-9y+9xy'}{y^2}=(-9) \cdot \frac{y-xy'}{y^2}= (-9) \cdot \frac{y-(-9)\frac{x^2}{y}}{y^2}$$
$$ \quad =(-9) \cdot \frac{y+9\frac{x^2}{y}}{y^2} = (-9) \cdot \frac{y^2+9x^2}{y^3}$$

 

[quicksilver123]

All I see in your post is "math processing error"

 

[fresh_42]

All I see in your post is "math processing error"

Hmmm, not in mine.

Sorry, I tried to upload it as an image, but it did not work.

Perhaps a repetition works.

This sign error is pretty resistant. May have to do with your linear notation.
You have $y''=(-9y-(-9x)y')/y^2$ which is
$$y''=\frac{-9y-(-9x)y'}{y^2}=\frac{-9y+9xy'}{y^2}=(-9) \cdot \frac{y-xy'}{y^2}= (-9) \cdot \frac{y-(-9)\frac{x^2}{y}}{y^2}$$
$$ \quad =(-9) \cdot \frac{y+9\frac{x^2}{y}}{y^2} = (-9) \cdot \frac{y^2+9x^2}{y^3}$$

If not, here is the source code:

You have y''=(-9y-(-9x)y')/y^2 which is
y''=\frac{-9y-(-9x)y'}{y^2}=\frac{-9y+9xy'}{y^2}=(-9) \cdot \frac{y-xy'}{y^2}= (-9) \cdot \frac{y-(-9)\frac{x^2}{y}}{y^2}
\quad =(-9) \cdot \frac{y+9\frac{x^2}{y}}{y^2} = (-9) \cdot \frac{y^2+9x^2}{y^3}

 

[quicksilver123]

Thanks, but it still looks nothing like the answer given in the book.. why is that?

 

[fresh_42]

Thanks, but it still looks nothing like the answer given in the book.. why is that?

Because I left out the final step where the knowledge about the value of $9x^2 + y^2$
(9x^2 + y^2)
has to be taken into account.

 

[quicksilver123]

...

 

[fresh_42]

By definition $9x^2 + y^2 = 9$
So $y'' = \dots = (-9) \cdot \frac{y^2+9x^2}{y^3}= (-9) \cdot \frac{9}{y^3}= -81 \frac{1}{y^3}$.

By definition 9x^2 + y^2 = 9.
So y'' = \dots = (-9) \cdot \frac{y^2+9x^2}{y^3}= (-9) \cdot \frac{9}{y^3}= -81 \frac{1}{y^3}.

 

[quicksilver123]

Thanks.