I am not sure if I got the correct answer or not. It was a homework from my textbook, and it does not have answers for even number questions...

Original Function: $x^2y-4x=5$
$\frac{dy}{dx}$ = $\frac{4-2xy}{x^2}$

It seems I got to finding the dy/dx part correctly according to wolfram alpha
Heres the steps to finding $\frac{dy^2}{d^2x}$

  1. $\frac{(x^2)(-2y-2x\frac{dy}{dx}) - ((-2xy+4)(2x))}{x^4}$
  2. $\frac{(x^2)(-2y-2x\frac{4-2xy}{x^2}) - ((-2xy+4)(2x))}{x^4}$
  3. $\frac{(-2x^2y+4x^2y-8x) - (-4x^2y+8x)}{x^4}$
  4. $\frac{6x^2y-16x}{x^4}$
  5. $\frac{6(5+4x)-16x}{x^4}$
  6. $\frac{30+24x-16x}{x^4}$
  7. $\frac{30+8x}{x^4}$ = $\frac{dy^2}{d^2x}$

Can anyone tell me if I did it correctly?

  • $\begingroup$ It's a lot less messy doing it the standard way. Your answer appears to be correct. $\endgroup$ – Mike Sep 27 '14 at 19:12
  • $\begingroup$ what's the standard way? I am working on these problems on pieces of paper, and it takes so much of the space to solve them... $\endgroup$ – hs2345 Sep 27 '14 at 19:15
  • $\begingroup$ Why not to simply use the chain rule and express $y''$ as a function of $y'$, $y$ and $x$ ? $\endgroup$ – Claude Leibovici Sep 27 '14 at 19:22
  • $\begingroup$ I mean without implicit differentiation, you have $y=4x^{-1}+5x^{-2}$. It's easy to see if you differentiate this twice, you'll get what you have calculated. $\endgroup$ – Mike Sep 27 '14 at 19:30


You properly obtained by implicit differentiation $$y'(x)=\frac{4-2xy(x)}{x^2}$$ So, take the derivative wrt $x$ and obtain $$y''(x)=\frac{-2 x y'(x)-2 y(x)}{x^2}-\frac{2 (4-2 x y(x))}{x^3}=\frac{2 x \left(y(x)-x y'(x)\right)-8}{x^3}$$ Now, if you want to plug $y'(x)$, just do it.

If I may suggest, remember that it is almost always the first derivative which can be difficult to obtain. Later, chain rule, product and quotient rules are your best friends.

  • $\begingroup$ You are very welcome ! $\endgroup$ – Claude Leibovici Sep 27 '14 at 19:34

i have also $y'=\frac{4-2xy}{x^2}$ and by the quotient rule $y''=\frac{(-2y-2xy')x^2-(4-2xy)2x}{x^4}$ plugging $y''$ in the equation above we obtain $y''=\frac{\left(-2y-2x\left(\frac{4-2xy}{x^2}\right)\right)x^2-(4-2xy)2x}{x^4}$ simplifying this we get $y''=\frac{6xy-16}{x^3}$

  • $\begingroup$ Oh... so I can stop there, and there is no need to plug in the original function into the terms of x and y? $\endgroup$ – hs2345 Sep 27 '14 at 19:30
  • $\begingroup$ yes you can stop there $y''$ is a function in $x$ and $y(x)$ $\endgroup$ – Dr. Sonnhard Graubner Sep 27 '14 at 19:31

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