How to take the third derivative of this function without it getting extremely messy?

Question

I have the equation $x^2 +xy + y^3 = 1$ and I'm being asked to find $y'''(x)$ at the point where $x=1$.

I calculate $f'(x) = {{-y-2x}\over{x+3y^2}}$

When I try to take the second derivative of the function, things get extremely messy, making it practically impossible. What can I do? I feel like I am missing something. I even tried logarithmic differentiation.

Answer 1

You can differentiate implicitly:

\begin{align} x^2 + xy + y^3 &= 1 &\big| \frac{d}{dx} \\ 2x + y + xy' + 3y^2 y' &= 0 &\big| \frac{d}{dx} \\ 2 + y' + y' + xy'' + 6y (y')^2 + 3y^2y'' &= 0 &\big| \frac{d}{dx} \\ 2y'' + y'' + xy''' + 6(y')^3 + 12 y y' y'' + 6yy'y'' + 3y^2y''' &= 0 \end{align}

The resulting equation is linear in $y'''$ an can be solved easily.

Answer 2

Use implicit differentiation, and then solve for $y$. Just take the derivative of both sides:

$$2x + x y'(x) + y(x)+ 2(y(x))^2y'(x)=0$$

You can then calculate then calculate de next derivatives and try to solve for $y'''(x)$.

Answer 3

We can proceed in much the same way as when we find the first derivative: by implicit differentiation. The difference is that we will have a system of equations to solve.

Differentiating once gives us: $$2x+y+xy'+3y^2y'=0.\tag1$$

Differentiating again gives us: $$2+2y'+xy''+6y(y')^2+3y^2y''=0.\tag2$$

One more differentiation gives us: $$3y''+xy'''+6(y')^3+18yy'y''+3y^2y'''=0\tag3$$

To make things easier, remember that we only care about the $x=1$ case! Thus, we're really only interested in solving the system: $$2+a+b+3a^2b=0\tag4$$ $$2+2b+c+6ab^2+3a^2c=0\tag5$$ $$3c+d+b^3+18abc+3a^2d\tag6$$

Here, $a=y(1),$ $b=y'(1),$ $c=y''(1),$ $d=y'''(1).$ Then we first solve $(4)$ for $b,$ use that result to solve $(5)$ for $c,$ and use those two results to solve for $d.$ Finally, we perform the substitutions $a\mapsto y(1)$ and $d\mapsto y'''(1),$ and we're done.

It's still not overly nice, and there are a bunch of calculations to check, but this makes the process a bit less messy.

Answer 4

At $x = 1$, we have

$$1 + y(1) + y(1)^3 = 1 \iff y(1)(1 + y(1)^2) = 0 \implies y(1) = 0$$

Rewrite the equation as

$$x^2 + xy + y^3 = 1 \quad\iff\quad y(x) = \frac{1}{x} - x - \frac{y(x)^3}{x}\tag{*1}$$

Differentiate the equation on RHS of $(*1)$, we get

$$y'(x) = -\frac{1}{x^2} - 1 - \frac{d}{dx}\left[\frac{y(x)^3}{x}\right]$$

If one expand the rightmost factor out, all terms of it contains a factor $y(x)$. This means its contribution vanish at $x = 1$. This leads to

$$y'(1) = -\frac{1}{1^2} - 1 = -2$$

Differentiate the equation on RHS of $(*1)$ three times, we get

$$y'''(x) = -\frac{3!}{x^4} - \frac{d^3}{dx^3}\left[\frac{y(x)^3}{x}\right]$$

We can evaluate the rightmost factor by General Leibniz rule.

$$\frac{d^3}{dx^3}\left[\frac{y(x)^3}{x}\right] = \sum_{\substack{p, q, r, s \ge 0 \\ p + q + r + s = 3}} \frac{3!}{p!q!r!s!} \frac{d^p x^{-1}}{dx^p} \frac{d^q y(x)}{dx^q} \frac{d^r y(x)}{dx^r} \frac{d^s y(x)}{dx^s} $$ Since $y(1) = 0$, the terms that contribute at $x = 1$ need to satisfy $q, r, s > 0$.
Since $p, q, r, s \ge 0$ and $p + q + r + s = 3$, this forces $p = 0, q = r = s = 1$.
This means at $x = 1$, we have

$$y'''(1) = -\frac{6}{1^4} - \frac{3!}{0!(1!)^3} \frac{y'(1)^3}{1} = -6(-2)^3 - 6 = 42$$