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Hello guys I want to make sure myself in determming when is $y$ as function of $x$,so for this, let us consider following question. If the equation of circle is given by this

$$x^2+y^2=25$$ and question is find the equation of tangent of circle at point $(3,4)$,then it is clear that we need calculate the derivative and also we can express $ y$ as

$$y=\sqrt{25-x^2}$$ but, let us consider following situation

$$x^3+y^3=6\cdot x \cdot y$$

it's name is The Folium of Descartes and my question is the same : calculate the equation of the tangent of this folium at point $(3,4)$. I know that somehow expressing $y$ as a function of $x$ is difficult but could't we do it by some even long mathematical manipulation? So what is the strict explanation when is $y$ a function of $x$ and when is it not? I need to be educated in this topic and please explain me ways of this,thanks a lot.

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up vote 4 down vote accepted

The equation $$ \tag{1}x^3+y^3=6xy $$does define $y$ as a function of $x$ locally (or, rather, it defines $y$ as a function of $x$ implicitly). Here, it is difficult to write the defining equation as $y$ in terms of $x$. But, you don't have to do that to evaluate the value of the derivative of $y$.

[edit] The point $(3,4)$ does not satisfy equation (1); so there is no tangent line at this point.

Let's, instead, consider the point $(3,3)$, which does satisfy equation (1):

To find the slope of the tangent line at $(3,3)$, you need to find $y'(3)$. To find this, first implicitly differentiate both sides of the defining equation for $y$ (equation (1)). This gives $$ {d\over dx} (x^3+y^3)={d\over dx} 6xy $$ So, using the chain and product rules: $$ 3x^2+3y^2 y' =6y+6x y'. $$ When $x=3$ and $y=3$, you have $$ 3\cdot 3^2+3\cdot 3^2 y'(3)=6\cdot 3+6\cdot 3\cdot y'(3). $$ Solve this for $y'(3)={3\cdot 3^2-6\cdot 3\over6\cdot3 -3\cdot3^2}=-1.$

Now you can find the equation of the tangent line since you know the slope and that the point $(3,3)$ is on the line.

Generally any "nice" equation in the variables $x$ and $y$ will define $y$ as a function of $x$ in some neighborhood of a given point. Given the $x$ value, the corresponding $y$ value is the solution to the equation ("the" solution in a, perhaps small, neighborhood of the point).

Of course, sometimes it is extremely difficult (if not impossible) to find an explicit form of the function; that is, of the form $y=\Phi(x)$ for some expression $\Phi(x)$. In these cases, to find the derivative of $y$, you have to use the approach above.

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Just to make sure that the OP won't jump to wrong conclusions: The above method works exactly, when we don't need to divide by zero when solving the derivative $y'$ from the above equation. The implicit function theorem then tells exactly what is needed "locally" $y$ can be solved uniquely as a function of $x$. But things may go wrong. If you try this with the folium of Descartes, when $(x,y)=(0,0)$, you will not succeed. The reason is that the folium intersects itself there, and there cannot be a (unique) tangent. You can isolate the two branches with other tricks. – Jyrki Lahtonen Dec 15 '11 at 15:23

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