Implicit differentiation.

Sometimes a graph in the plane is defined by an equation of the form $ F(x,y)=0$ , but this equation cannot be put into the form $ y=f(x)$ . In such a case we cannot compute a derivative $ \frac {dy}{dx}$ by the ordinary methods. Nevertheless, on some well-chosen intervals, it is possible to do it by implicit differentiation. We will see that on examples.

Example 6.15.1   Let $ \mathcal{E}$ be the ellipse whose equation is $ \frac {x^2}{9} + \frac {y^2}{4} =1$ (displayed on Figure 17. At the vertices $ A$ and $ A'$ , it has tangents parallel to the $ y-$ axis, therefore these tangents have no slope. But at any other point the tangent is well-defined; the reason is the following: the graph can be written as $ \mathcal{E}=\mathcal{E_1} \cup \mathcal{E_2}$ , where $ \mathcal{E_1}$ is the upper half ellipse (whose equation is $ y= f(x)=\frac 23 \sqrt{9-x^2}$ and $ \mathcal{E_2}$ is the lower half ellipse (whose equation is $ y= -f(x)=-\frac 23 \sqrt{9-x^2}$ . For $ x \neq \pm 3$ , the function $ f$ is differentiable and, at the corresponding point on the graph, there is a tangent which has a well-defined slope.

Figure 17: Tangents to an ellipse.
\begin{figure}\mbox{\epsfig{file=ImplicitEllipse.eps,height=4cm}}\end{figure}

We will not do it here in this way. Consider the equation of $ \mathcal{E}$ :

$\displaystyle \frac {x^2}{9} + \frac {y^2}{4} =1$    

We differentiate each side with respect to $ x$ :

\begin{equation*}\begin{aligned}\frac {d}{dx} \left( \frac {x^2}{9} + \frac {y^2...
...} & = 0  \frac {2x}{9} + \frac 12 y \cdot y' & =0 \end{aligned}\end{equation*}

If $ y \neq 0$ , we have:

$\displaystyle y' = -\frac {4x}{9y}.$    

E.g. the slope of the tangent to $ \mathcal{E}$ at the point whose coordinates are $ (2, 2\sqrt{5} / 3 )$ is equal to $ \frac{-4 \sqrt{5}}{3}$ .

Example 6.15.2   Let $ \mathcal{C}$ be the curve with equation $ x^2+y^2-xy=3$ . By implicit differentiation with respect to $ x$ we have:

\begin{equation*}\begin{aligned}\frac {d}{dx} \left( x^2+y^2-xy \right) & = \fra...
...t y' - y - x \cdot y' &= 0  (2y-x) y' +2x -y & =0 \end{aligned}\end{equation*}

At every point such that $ 2y-x \neq 0$ , $ y$ is a differentiable fuction of $ x$ and we have:

$\displaystyle y' = \frac {2x-y}{x-2y}$    

E.g., the slope of the tangent to $ \mathcal{E}$ at the point whose coordinates are $ (1,2)$ is equal to 0.

Noah Dana-Picard 2007-12-28