# implicit differentiation

An equation is defined as $x+y+x^5-y^5=0$.

$a$. Show that the equation determines $y$ as a function of $x$ in a neighbourhood of the origin $(0,0)$.

$b$. Denote the function from $(a)$ by $y=\varphi(x)$. Find $\varphi^{(5)}(0)$ and $\varphi^{(2004)}(0)$.

For the first part,I simply needed to prove that the equation is $C^1$ in the neighbourhood of the origin and that the first derivative w.r.t.y does not vanish at that point. These are the sufficient conditions for the existence of an implicit function in the neighbourhood of the origin.

For the second part, I could do implicit differentiation (in a usual way) had it been for lower orders. But I need to learn the case for higher orders like the question asks. Please give me some hints.

After one derivative, you have $$1 + y' + 5x^4 - 5y^4y' = 0.$$ Now, to find higher derivatives, make sure that you think about the chain rule. The implicit second derivative equation is $$y'' + 20x^3 - 20y^3(y')^2 - 5y^4y'' = 0.$$ The third derivative equation: $$y''' + 60x^2 - 60y^2(y')^3 - 40y^3y'y'' - 20y^3y'y'' - 5y^4y''' = 0.$$ Notice how several of the terms have a polynomial factor. With each derivative, the degree decreases. Can you see what happens after $5$ derivatives? In the long run?

• I have some vague idea but I cannot comprehend completely. Perhaps if you could give me the full solution, I might see the pattern. – user77440 May 12 '13 at 7:55
• You must try to do it yourself! You can edit your question to show what you got, and we can help you check it. – Sammy Black May 12 '13 at 8:01
• Sammy, I have been trying for a long time now. I get that $\varphi'(0)$ =-1. Do I need to look for some sort symmetry or what? If yes, please tell me how. – user77440 May 12 '13 at 11:45

What makes the second question seem intimidating is that we are used to being asked to find specific expressions for the various higher derivatives found by implicit differentiation, but that isn't called for here.

Sammy Black shows the start of the "chain" of implicit differentiations that we need to make on the equation of the curve. There are a couple of things we can do to make the resolution of the problem clearer. One is that we do not need to do even so much as to collect the terms in the derivative equations, because of the second item. Since we are asked about the behavior of the higher derivatives of the implicit function $\ y \ = \ \varphi ( x ) \$ in the neighborhood of the origin , the amount of effort we need to make is reduced immensely.

The first round of differentiation (as already shown) produces

$$\frac{d}{dx} \ [ \ x \ + \ y \ + \ x^5 \ - \ y^5 \ ] \ = \ \frac{d}{dx} [ \ 0 \ ] \ \ \Rightarrow \ \ 1 \ + \ y \ ' \ + \ 5x^4 \ - \ 5y^4 \ y \ ' \ \ = \ 0$$

$$\Rightarrow \ \ 1 \ + \ 5x^4 \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ ' \ \ = \ 0 \ \ .$$

We wish to evaluate the left-hand side at $\ (x,y) \ = \ (0,0) \$ , which gives us

$$\Rightarrow \ \ 1 \ + \ 5 \cdot 0 \ + \ ( \ 1 \ - \ 5 \cdot 0 \ ) \ y \ ' \ \ = \ 0 \ \ \Rightarrow \ \ y \ ' \ = \ \varphi \ ' (0) \ = \ -1 \ \ .$$

We next implicitly differentiate our first-derivative equation repeatedly as it stands, and perform a similar evaluation each time:

$$\frac{d}{dx} \ [ \ 1 \ + \ 5x^4 \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ ' \ ] \ = \ \frac{d}{dx} [ \ 0 \ ]$$

$$\Rightarrow \ \ 20x^3 \ + \ ( \ -20y^3 \cdot y \ ' \ ) \cdot y \ ' \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ '' \ \ = \ 0$$

$$\Rightarrow \ \ \text{at (0,0)} : \ \ 20 \cdot 0 \ + \ ( \ 20 \cdot 0 \cdot y \ ' \ ) \cdot y \ ' \ + \ ( \ 1 \ - \ 5 \cdot 0 \ ) \ y \ '' \ \ = \ 0$$

$$\Rightarrow \ \ y \ '' \ = \ \varphi \ '' (0) \ = \ 0 \ \ ;$$

$$\frac{d}{dx} \ [ \ 20x^3 \ + \ ( \ -20y^3 \ ) \cdot ( \ y \ ' \ )^2 \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ '' \ ] \ = \ \frac{d}{dx} [ \ 0 \ ]$$

$$\Rightarrow \ \ 60x^2 \ + \ ( \ -60y^2 \cdot y \ ' \ ) \cdot ( \ y \ ' \ )^2 \ + \ ( \ -20y^3 \ ) \cdot ( \ 2 y \ ' \ y \ '' \ ) \ + \ ( \ - 20y^3 \cdot y \ ' \ ) \ y \ '' \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y \ ''' \ \ = \ 0$$

$$\Rightarrow \ \ \text{at (0,0)} : \ \ 60 \cdot 0 \ - \ 60 \cdot 0 \cdot 0 \ - \ 20 \cdot 0 \cdot 0 \ - \ 20 \cdot 0 \cdot 0 \ + \ ( \ 1 \ - \ 5 \cdot 0 \ ) \ y \ ''' \ \ = \ 0$$

$$\Rightarrow \ \ y \ ''' \ = \ \varphi \ ''' (0) \ = \ 0 \ \ .$$

At this point, we notice something that will relieve the tedium of computing all those products of higher derivatives: except for the first and last terms on the left-hand side, all the others involve only $\ y \$ and its derivatives, all of which so far have proven to equal zero. So for the successive higher derivatives of the left-hand side of the equation, we can just focus on the terms that behave otherwise:

fourth derivative --

$$120x \ + \ \ldots \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y^{(4)} \ \ = \ 0 \ \ \Rightarrow \ \ y^{(4)} \ = \ \varphi^{(4)} (0) \ = \ 0 \ \$$

[still nothing interesting, but we can see what's coming...]

fifth derivative --

$$120 \ + \ \ldots \ + \ ( \ 1 \ - \ 5y^4 \ ) \ y^{(5)} \ \ = \ 0 \ \ \Rightarrow \ \ y^{(5)} \ = \ \varphi^{(5)} (0) \ = \ -120 \ \ \text{(!)}$$

We at last reach a result that perhaps ends our wondering why the implicit function isn't simply $\ y \ = \ -x \$ . What we learn from this is that the curve described by our equation is extremely flat in the neighborhood of the origin:  in this enlargement, the red line ( y = -x ) indicates where our curve begins to deviate significantly from a straight line

As for still higher derivatives, it is now apparent that, with the term involving $\ x \$ "differentiated away", the only non-zero derivatives of $\ y \$ will only be found in terms which are products with other factors which equal zero. We may immediately conclude from this that $\ \varphi^{(k)} (0) \ = \ 0 \$ , for $\ k \ \ge \ 6 \ .$

[Note that this generalizes easily for equations $\ x \ + \ y \ + \ x^n \ - \ y^n \ = \ 0 \$ , for odd $\ n \$ . When $\ n \$ is even, the equation describes the union of a curve which does not pass through the origin and the "degenerate" line $\ y \ = \ -x \$ , for which all derivatives beyond $\ y \ ' \$ are zero at the origin.]