Implicit Differentiation

Section 2.6 Implicit Differentiation

Sometimes we have an equation that can't easily be solved for

y,

for example:

x^{3} + 3 y^{3} = 3 x y .

It helps to consider

y = y (x) .

With that consideration, it might be possible to calculate

\frac{d y}{d x} .

Differentiate

x^{3} + 3 y^{3} = 3 x y :

\begin{aligned} \frac{d}{d x} (x^{3} + 3 y^{3}) & = \frac{d}{d x} (3 x y) \\ \frac{d}{d x} (x^{3}) + \frac{d}{d x} (3 y^{3}) & = 3 x \cdot \frac{d y}{d x} + y \cdot \frac{d}{d x} (3 x) & (separate the terms) \\ 3 x^{2} + 3 \cdot 3 y^{2} \frac{d y}{d x} & = 3 x \cdot \frac{d y}{d x} + y \cdot 3 & (apply the chain rule) \\ 3 x^{2} + 9 y^{2} \frac{d y}{d x} & = 3 x \frac{d y}{d x} + 3 y & (solve for \frac{d y}{d x}) \\ 9 y^{2} \frac{d y}{d x} - 3 x \frac{d y}{d x} & = 3 y - 3 x \\ \frac{d y}{d x} (9 y^{2} - 3 x) & = 3 y - 3 x \\ \frac{d y}{d x} & = \frac{3 y - 3 x}{9 y^{2} - 3 x} \end{aligned}

Find the slope of the tangent to $x^{2} + y^{2} = 1,$ at the point $(\frac{1}{2}, \frac{\sqrt{3}}{2}) :$

\begin{aligned} x^{2} + (y (x))^{2} & = 1 & (treat y as y(x)) \\ \frac{d}{d x} (x^{2} + (y (x))^{2}) & = \frac{d}{d x} (1) \\ 2 x + 2 y \cdot \frac{d y}{d x} & = 0 \\ 2 y \cdot \frac{d y}{d x} & = - 2 x \\ \frac{d y}{d x} & = - \frac{x}{y} \end{aligned}

At $(\frac{1}{2}, \frac{\sqrt{3}}{2}),$ $M_{tan} = \frac{- \frac{1}{2}}{\frac{\sqrt{3}}{2}} = - \frac{1}{\sqrt{3}}$

Figure 2.6.1.