I have a problem to solve:

use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^2y}{dx^2}$. Write the solutions in terms of x and y only

It means that I need to differentiate the equation one time to find $y'$ and then once more to find $y''$.

The correct answer from the textbook is $y' = \frac{x + 1}{y}$ and $y'' = \frac{x^2 + 2x}{y^3}$. I got the first derivative right, but I can't understand how did they get the second one, or is it a typo (unlikely), since I have $y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3}$

I did this:

$$ y^2 = x^2 + 2x\\ 2yy' = 2x + 2\\ yy' = x + 1\\ y' = \frac{x + 1}{y}\\ $$

I tried to get to the second derivative from both $yy' = x + 1$, $y' = \frac{x + 1}{y}$ and $2yy' = 2x + 2$. But every time I had that dangling constant (1 or 2), which lead to the dangling $\frac{1}{y}$ in my answer.

Like here:

$$ yy' = x + 1\\ y'y' + yy'' = 1\\ yy'' = 1 - (y')^2\\ y'' = \frac{1 - (y')^2}{y}\\ y'' = \frac{1}{y} - \frac{(y')^2}{y}\\ y'' = \frac{1}{y} - \frac{(\frac{x + 1}{y})^2}{y}\\ y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3} $$

I don't see any way to get from my answer to the textbook's one with a transformation, no way to get rid from y in the numerator. And the correct answer doesn't have a "y" there.

Could someone either point to an error in my solution, or corroborate the suspicion that it indeed may be a typo.

  • 5
    $\begingroup$ I think it is a typo. Recall that you have $y^2=x^2+2x$, so $$ y'' = \frac{y^2 - (x+1)^2}{y^3} = \frac{2x+x^2 - (x+1)^2}{y^3} = \frac{-1}{y^3}. $$ $\endgroup$ – Math1000 Nov 16 '19 at 21:00
  • $\begingroup$ this must be the first typo I encountered in that textbook in 200 pages, which looks like a very low amount, I would say the textbook is of exceptional quality $\endgroup$ – user907860 Nov 16 '19 at 21:05
  • $\begingroup$ What is the textbook? Perhaps there is a list of errata online somewhere. $\endgroup$ – Math1000 Nov 16 '19 at 21:06
  • $\begingroup$ thomas' calculus 14th edition $\endgroup$ – user907860 Nov 16 '19 at 21:06
  • 1
    $\begingroup$ As shown in @zwim's answer, the correct result is $y' = \frac{-1}{y^3}$. You could always email the textbook's author with a link to this page and they would likely fix it for the 15th edition. $\endgroup$ – Math1000 Nov 16 '19 at 21:16

From $y'y'+yy''=1$ multiply by $y^2$.

Then $(yy')^2+y^3y''=(x+1)^2+y^3y''=y^2=x^2+2x \iff y^3y''=x^2+2x-x^2-2x-1=-1$

If we continue your calculation

$y''=\dfrac 1y-\dfrac{(x+1)^2}{y^3}=\dfrac{y^2-(x+1)^2}{y^3}=\dfrac{(x^2+2x)-(x^2+2x+1)}{y^3}=\dfrac{-1}{y^3}$

Gives the same result, so I guess the textbook result is erroneous (i.e. it gives $yy''=1$ which does not agree with derivatives of $\pm\sqrt{x^2+2x}$)


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