I have actually the equation $x^2 +xy + y^3 = 1$ and also I"m gift asked to discover $y"""(x)$ in ~ the suggest where $x=1$.

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I calculate $f"(x) = -y-2x\overx+3y^2$

When I try to take it the second derivative of the function, things get extremely messy, make it virtually impossible. What can I do? ns feel favor I am absent something. I also tried logarithmic differentiation. You can identify implicitly:

\beginalignx^2 + xy + y^3 &= 1&\big| \fracddx \\2x + y + xy" + 3y^2 y" &= 0&\big| \fracddx \\2 + y" + y" + xy"" + 6y (y")^2 + 3y^2y"" &= 0 &\big| \fracddx \\2y"" + y"" + xy""" + 6(y")^3 + 12 y y" y"" + 6yy"y"" + 3y^2y""" &= 0\endalign

The resulting equation is linear in $y"""$ an deserve to be solved easily. Use implicitly differentiation, and then deal with for $y$. Just take the derivative the both sides:

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

You can then calculation then calculate de following derivatives and shot to fix for $y"""(x)$. We deserve to proceed in lot the same way as as soon as we discover the an initial derivative: by implicit differentiation. The difference is that us will have actually a system the equations to solve.

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

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

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

To make points easier, remember the we only care about the $x=1$ case! Thus, we"re really just interested in fixing 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, $a=y(1),$ $b=y"(1),$ $c=y""(1),$ $d=y"""(1).$ climate we an initial solve $(4)$ for $b,$ usage that result to deal with $(5)$ for $c,$ and also use those two results to resolve for $d.$ Finally, we carry out 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 come check, however this renders the procedure a bit much less messy.

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answered Oct 20 "15 at 22:39 Cameron BuieCameron Buie
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At $x = 1$, us have

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

Rewrite the equation as

$$x^2 + xy + y^3 = 1 \quad\iff\quad y(x) = \frac1x - x - \fracy(x)^3x\tag*1$$

Differentiate the equation on RHS that $(*1)$, us get

$$y"(x) = -\frac1x^2 - 1 - \fracddx\left<\fracy(x)^3x\right>$$

If one increase the rightmost aspect out, all terms of it has a aspect $y(x)$.This way its donation vanish at $x = 1$. This leads to

$$y"(1) = -\frac11^2 - 1 = -2$$

Differentiate the equation ~ above RHS the $(*1)$ three times, us get

$$y"""(x) = -\frac3!x^4 - \fracd^3dx^3\left<\fracy(x)^3x\right>$$

We can evaluate the rightmost variable by general Leibniz rule.

$$\fracd^3dx^3\left<\fracy(x)^3x\right>= \sum_\substackp, q, r, s \ge 0 \\ ns + q + r + s = 3\frac3!p!q!r!s!\fracd^p x^-1dx^p\fracd^q y(x)dx^q\fracd^r y(x)dx^r\fracd^s y(x)dx^s$$Since $y(1) = 0$, the state that add at $x = 1$ need to meet $q, r, s > 0$.Since $p, q, r, s \ge 0$ and also $p + q + r + s = 3$, this pressures $p = 0, q = r = s = 1$.This means at $x = 1$, us have

$$y"""(1) = -\frac61^4 - \frac3!0!(1!)^3 \fracy"(1)^31 = -6(-2)^3 - 6 = 42$$