Find $y''$ if $x^4 + y^4 = 16$ by implicit differentiation

So after the first implicit differentiation I got this equation (let's call it A):

$4x^3 + 4y^3*\frac{dy}{dx} = 0$ Where $\frac{dy}{dx}$ is $y'$

At this point the text book finds the second derivative by making $\frac{dy}{dx}$ the subject and getting a value of $\frac{dy}{dx}$ in terms of y and x which is $\frac{-x^3}{y^3}$ and then taking $\frac{d}{dx}$ of $\frac{-x^3}{y^3}$ to work out $\frac{d^2y}{dx^2}$. I understand that there is nothing wrong with this approach. However I'm wondering why we can't do implicit differentiation again on Equation A(written above) just like we did on the original equation, in order to find $y''$. My reasoning is given below, please tell me what's wrong with it.

$\frac{d}{dx}(4x^3 + 4y^3*\frac{dy}{dx}=0)$

$12x^2 + [12y^2\frac{dy}{dx}+\frac{d^2y}{dx^2}4x^3] = 0$

Making $\frac{d^2y}{dx^2}$ the subject I get:


However the textbook gives the answer as:

$\frac{-3x^2(x^4+y^4)}{y^7}$ and since $x^4+y^4=16$

$\frac{-3x^2(16)}{y^7}$ = $\frac{-48x^2}{y^7}$

I feel like there's a missing algebraic link between the textbook's answer and mine, or maybe my second order differentiation was wrong? Let me know, thank you.


Your approach is valid, but it's easy to lose ourselves in the notation. I will make the substitution $g = \frac{dy}{dx}$.

Equation $(A)$ becomes $4x^3 + 4y^3 g = 0$. Differentiating with respect to $y$, we have

$$12x^3 + 12y^2\frac{dy}{dx}g+4y^3\frac{dg}{dx}=0.$$

Notice that another $\frac{dy}{dx}$ popped up from $\frac{d}{dx}4y^3$. Now, $$12x^3 + 12y^2\left(\frac{dy}{dx}\right)^2+4y^3\frac{d^2y}{dx^2}=0.$$

Solving this for $\frac{d^2y}{dx^2}$ will yield the correct result after some manipulation.

  • $\begingroup$ Ah! Got it, thanks! I missed one of the $\frac{dy}{dx}$ during the implicit differentiation on equation A. Just two remarks: shouldn't the $12x^3$ be $12x^2$? Also, it is correct to say differentiating with respect to $y$? I thought since we were performing implicit differentiation with $\frac{d}{dx}$ on equation A it was called differentiating with respect to $x$. Correct me if I am wrong. $\endgroup$ – Eric Nov 8 '14 at 10:49

The more specific reason is - you appeared to have forgotten to use the chain rule to solve for the $4y^3$ part of $4y^{3}\frac{dy}{dx}$

You incorrectly solved as follows:

$\frac{d}{dx}[4y^3] \neq 12y^2$

Rather than:

$\frac{d}{dx}[4y^3] = 12y^2\frac{dy}{dx}$

Which you didn't notice because of another $\frac{dy}{dx}$ which did appear in your chain rule. i.e.

$u=4y^3$ and $v=\frac{dy}{dy}$


$u'=12y^2\frac{dy}{dx}$ $v'=\frac{d^2y}{d^x}$

...therefore... $(uw)'=[12y^2\frac{dy}{dx}\frac{dy}{dy}+\frac{d^2y}{dx^2}4y^3]$

i.e. You missed one $\frac{dy}{dx}$ and got:



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