Differentiation of implicit functions

When $y$ is defined as an implicit function of $x$ by means of an equation in the form

$$f(x,y) = 0,$$ (5.4)

it was explained in the last section how it might be inconvenient to solve for $y$ in terms of $x$; that is, to find $y$ as an explicit function of $x$ so that the formulas we have deduced in this chapter may be applied directly. Such, for instance, would be the case for the equation

$$ax^6 + 2x^3y - y^7x - 10 = 0.$$ (5.5)

We then follow the rule:

Differentiate, regarding $y$ as a function of $x$, and put the result equal to zero ^5.13. That is,

$$\frac{d}{dx} f(x, y) = 0.$$ (5.6)

Let us apply this rule in finding $\frac{dy}{dx}$ from (5.5): by (5.6),

$$\frac{d}{dx}(ax^6 + 2x^3y - y^7x - 10) = 0,$$

$$\frac{d}{dx}(ax^6) + \frac{d}{dx}(2x^3y) - \frac{d}{dx}(y^7x) - \frac{d}{dx}(10) = 0;$$

$$6ax^5 + 2x^3 \frac{dy}{dx} + 6x^2y - y^7 - 7xy^6\frac{dy}{dx} = 0;$$

$$(2x^3 - 7xy^6)\frac{dy}{dx} = y^7 - 6ax^5 - 6x^2y;$$

$$\frac{dy}{dx} = \frac{y^7 - 6ax^5 - 6x^2y}{2x^3 - 7xy^6} .$$

This is the final answer.

The student should observe that in general the result will contain both $x$ and $y$.