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We have a curve $A$, which consists of all points $(x,y) \in \mathbb{R}^2$ which satisfy $$9x + 27y - \dfrac{10}{81} (x+y)^3 = 0$$

You're given that the curve $A$, sufficiently close to $(0,0)$, is the graph of an infinitely differentiable function $f$ (which is defined on a small, open interval which includes $0)$. Find $f"(0)$.

I don't know how to tackle this problem. My first step was to try and find $f'(x)$ obviously, but I don't know how to do it. I haven't studied multivariable calculus yet, so I'm not exactly sure how to find the tangent line to $A$ in $(0,0)$. However, intuitively I'd say you could also find a tangent line in the form of $y=ax+b$, so you could do it without the definition of the total derivative (which I googled), but now I don't know how to find the tangent line $y=ax+b$. Anyone have tips?

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    $\begingroup$ try implicit derivation: en.wikipedia.org/wiki/… $\endgroup$ – janmarqz Oct 7 '15 at 23:14
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    $\begingroup$ @janmarqz Why didn't you post the answer or some hints. $\endgroup$ – Arbuja Dec 12 '15 at 23:04
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    $\begingroup$ always it is worthwhile see the plot wolframalpha.com/input/… $\endgroup$ – janmarqz Dec 14 '15 at 3:51
  • $\begingroup$ @janmarqz Thanks for helping me fix my errors. $\endgroup$ – Arbuja Dec 15 '15 at 20:06
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Here's a way to get the answer quickly. We have

$$\tag 1 27f(x) = -9x +(x+f(x))^3.$$

for $x$ close to $0.$ Let $g(x)=(x+f(x))^3.$ Then $g\in C^\infty(\mathbb R).$ Because $f(0)=0,$ $f(x) = O(x)$ as $x\to 0.$ Hence $g(x)= O(x^3).$ A $C^\infty$ function that is $O(x^3)$ at $0$ has a vanishing second derivative at $0.$ Thus the right side of $(1)$ is a function whose second derivative is $0.$ Therefore $f''(0)=0.$

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  • $\begingroup$ Amazing! But I showed a general approach for all kinds of implicit curves $\endgroup$ – Arbuja Dec 15 '15 at 20:43
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HINT:

The implicit equation $F(x,y) = 0$ is given by an odd function ( $F(-x,-y) = - F(x,y)$) so $(-x,-y)$ is on the graph if $(x,y)$ is. Therefore, the explicit function $y = f(x)$ is odd, and so all the derivatives of even order at $0$ are $0$.

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  • $\begingroup$ Yeah a hint was probably better..... $\endgroup$ – Arbuja Dec 15 '15 at 22:01
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To find the first derivative you must take the derivative of the $x$'s and $y$'s and divide the $x$-derivative by the $y$-derivative to get $\frac{dy}{dx}$. Inorder for this to happen make sure you place $\frac{dy}{dx}$ next to the $y$ parts that is differentiated. (The tick marks represent the part of the original function that is differentiated).

$$9x+27y-\frac{10}{81}(x+y)^3=0$$ $$9x'+{27y}'-\left(\frac{10}{81}(x+y)^3\right)'=0$$ $$9+27\frac{dy}{dx}-\frac{30}{81}(x+y)^2\left(x'+y'\right)=0$$ $$9+\left(27\frac{dy}{dx}\right)-\frac{30}{81}(x+y)^2\left(1+\left(\frac{dy}{dx}\right)\right)=0$$

From here you should figure out..

$$\frac{dy}{dx}=\frac{-729+30(x+y)^{2}}{2187-30(x+y)^2}$$

Now to find the tangent line, at point $(a,b)$, know that

$y=f'(a,b)(x-a)+b$

Where $x=a$ and $y=b$.

Now for $(0,0)$ substitute $x=0$ and $y=0$ to the first derivative.

$$\frac{dy}{dx}=\frac{-729+30(x+y)^{2}}{2187-30(x+y)^2}=-\frac{729}{2187}=-\frac{1}{3}$$

$y=-\frac{1}{3}x$

However finding this tangent line is unhelpful for finding the answer to your problem. Instead you should continue to calculate the second derivative.

The second derivative is tricky but you should know that $\frac{dy}{dx}=-\frac{1}{3}$ so it can be calculated at $x=0$ more quickly.

Now to take the implict derivative of $f'(x)=\frac{-729+30(x+y)^{2}}{2187-30(x+y)^2}$ the second time.

For the sake of differentiating this quicker lets us not use the quotient rule instead lets convert the quotient to multiplication form.

$$\left({729+30(x+y)^{2}}\right)\left({2187-30(x+y)^2}\right)^{-1}$$

Now as long as you know the derivative multiplication and chain rules hopefully finding the second derivative is manageable for you. Remember $\left(f(x)\right)^{'}=f^{''}(x)=f^{2}(x)$. (The tick marks represents differentiation for certain parts of the first derivative.)

(1)$$\left(\frac{dy}{dx}\right)^{'}=\left({729+30(x+y)^{2}}\right)^{'}\left({2187-30(x+y)^2}\right)^{-1}+\left({729+30(x+y)^{2}}\right)\left(\left({\left({2187-30(x+y)^2}\right)^{-1}}\right)^{'}\right)$$ (2)$$\left(\frac{dy}{dx}\right)^{'}=\left({60(x+y)\left((x)^{'}+(y)^{'}\right)}\right)\left({2187-30(x+y)^2}\right)^{-1}-\left({729+30(x+y)^{2}}\right)\left({2187-30(x+y)^2}\right)^{-2}\left(-60(x+y)\left(1+\frac{dy}{dx}\right)\right)=$$ (3)$$\left(\frac{{dy}^2}{{dx}^2}\right)=\left({60(x+y)\left(1+\frac{dy}{dx}\right)}\right)\left({2187-30(x+y)^2}\right)^{-1}-\left({729+30(x+y)^{2}}\right)\left({2187-30(x+y)^2}\right)^{-2}\left(-60(x+y)\left(1+\frac{dy}{dx}\right)\right)=$$

(4) Now substituting $\frac{dy}{dx}=-1/3$ and $x=0,y=0$ $$\left(\frac{{dy}^2}{{dx}^2}\right)=\left({60(0)\left(\frac{2}{3}\right)}\right)\left({2187-30(0)^2}\right)^{-1}-\left({729+30(0)^{2}}\right)\left({2187-30(0)^2}\right)^{-2}\left(-60(0)\left(\frac{2}{3}\right)\right)=$$ (5)$$\left(\frac{{dy}^2}{{dx}^2}\right)=0$$ $$f''(0)=0$$ (P.S. This is an inflection point)

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  • $\begingroup$ Dear Down voter, why was this a bad answer. $\endgroup$ – Arbuja Dec 15 '15 at 22:29

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