I'm currently using the book University Calculus: Alternate Edition (Hass et al., 2008) to study calculus. I was studying implicit differentiation and ran into an exercise problem that I was having trouble solving. The question is:

Use implicit differentiation to find $dy/dx$ and $d^2y/dx^2$ for:

$$x^{2/3} + y^{2/3} = 1$$

I was able to find $y'$ fairly simply as:

$$y' = -\left( \dfrac{y}{x} \right)^{1/3}$$

but I'm having trouble finding $y''$.

My approach is as follows:

$$ \begin{align} y'' & = \frac{d}{dx}y' \\ & = \frac{d}{dx} \left( \frac{-y^{1/3}}{x^{1/3}} \right) \\ & = \frac{(-y^{1/3})'x^{1/3} - (-y^{1/3})(x^{1/3})'}{(x^{1/3})^2} \\ & = \frac{-\frac{1}{3}y^{-2/3} y' x^{1/3} + y^{1/3} \frac{1}{3}x^{-2/3}}{x^{2/3}} \\ & = \dfrac{\dfrac{-y^{-2/3} y' x^{1/3}}{3} + \dfrac{y^{1/3}}{3}}{x^{4/3}} \\ & = \dfrac{\left( -y^{-2/3} \times \dfrac{-y^{1/3}}{\phantom{-}x^{1/3}} \times x^{1/3}\right) + y^{1/3}}{3x^{4/3}} \\ & = \dfrac{y^{1/3} + y^{-1/3}}{3x^{4/3}} \end{align} $$

However, the textbook solution states that the correct answer for $y''$ is:

$$y'' = \dfrac{x^{2/3}y^{-1/3} - y^{1/3}}{3x^{4/3}}$$

and I'm having trouble reverse engineering where I went wrong. Any tips or pointers are appreciated. Thanks in advance.


In your fifth line, where you got

$$\dfrac{\dfrac{-y^{-2/3} y' x^{1/3}}{3} + \dfrac{y^{1/3}}{3}}{x^{4/3}} \tag{1}\label{eq1A}$$

you had multiplied by $x^{2/3}$ in the numerator and denominator, but you forgot to multiply the first term in the numerator by this factor. As such, it should have been

$$\dfrac{\dfrac{-y^{-2/3} y' x}{3} + \dfrac{y^{1/3}}{3}}{x^{4/3}} \tag{2}\label{eq2A}$$


Also, later you made a small sign error where both terms in the numerator should be positive as the $2$ negative signs cancel in the first term and the second term is already positive. Thus, with these $2$ changes, the final answer I get is

$$y'' = \dfrac{x^{2/3}y^{-1/3} + y^{1/3}}{3x^{4/3}} \tag{3}\label{eq3A}$$

It could be I made a mistake, there's a typo with the textbook solution, or you made a typo writing it in the question. You may wish to check on this.

  • 1
    $\begingroup$ Thanks for the answer and extra details. I double-checked the solution and the sign is a $-$... Perhaps it could be a typo, I've actually witnessed more of those than I ideally would like to. $\endgroup$
    – Sean
    Jan 20 '20 at 5:57
  • 1
    $\begingroup$ The negative sign mistake that I made is indeed a mistake. I'll edit in the correct terms. $\endgroup$
    – Sean
    Jan 20 '20 at 5:58
  • 1
    $\begingroup$ @Seankala You're welcome. I've checked the calculations again myself and I'm quite certain the official answer has a typo where it should be a $+$ instead. $\endgroup$ Jan 20 '20 at 6:03

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