Mathematics · Ch 5 — Continuity and Differentiability

Implicit Differentiation

So far every function differentiated has been given explicitly, in the form $y=f(x)$ , with

$y$ isolated on one side. Many relations between $x$ and $y$ , however, are given

implicitly -- as an equation $F(x,y)=0$ mixing both variables together, such as

$x^2+y^2=25$ or $x^3+y^3=3xy$ -- without $y$ ever being solved for explicitly (and sometimes it

cannot be, in terms of elementary functions at all). Implicit differentiation finds

$dy/dx$ directly from such an equation, without first solving for $y$ .

Method. Differentiate both sides of the equation with respect to $x$ , term by term, treating

$y$ throughout as an (unknown, but differentiable) function of $x$ . Every term containing $y$

must be differentiated using the chain rule (Section 3): for instance,

\frac{d}{dx}\big(y^2\big) = 2y\,\frac{dy}{dx}, \qquad \frac{d}{dx}\big(xy\big) = x\,\frac{dy}{dx}+y \quad\text{(product rule, since $y$ depends on $x$)}.

After differentiating, the resulting equation is linear in $dy/dx$ ; collect all terms containing

$dy/dx$ on one side and solve algebraically for it.

Worked illustration ( $x^2+y^2=25$ , a circle of radius $5$ ). Differentiating both sides with

respect to $x$ :

2x + 2y\,\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{x}{y}, \qquad y\neq0.

Geometrically, this says the tangent to a circle at any point $(x,y)$ (other than the top/bottom)

has slope $-x/y$ -- perpendicular to the radius to that point, as expected from circle geometry.

A second illustration ( $\sin(xy)=x$ ). Differentiating with the chain rule on the left (outer

function $\sin$ , inner function $u=xy$ , itself needing the product rule for $du/dx$ ):

\cos(xy)\cdot\left(x\frac{dy}{dx}+y\right) = 1 \quad\Longrightarrow\quad x\cos(xy)\frac{dy}{dx} = 1-y\cos(xy) \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1-y\cos(xy)}{x\cos(xy)}.

Why implicit differentiation is necessary, not just convenient. For an equation such as

$x^3+y^3=3xy$ (a folium-type curve), solving explicitly for $y$ in terms of $x$ is algebraically

impractical or impossible using elementary functions, yet the curve still has a well-defined

slope at each of its points. Implicit differentiation finds this slope directly from the

relation itself, sidestepping the need to solve for $y$ at all -- exactly the technique …