2
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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 $\pm\sqrt{x^2+2x}$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.