3.3 Implicit differentiation

Implicit Differentiation

Implicit differentiation lets you find derivatives of functions defined by equations you can't (or don't want to) solve for one variable explicitly. In single-variable calculus, you used this for curves like $x^2 + y^2 = 1$. In multivariable calculus, the same idea extends to functions of several variables, and partial derivatives plus the chain rule do the heavy lifting.

Implicit Differentiation

Differentiating Implicitly Defined Functions

An implicit function defines a relationship between variables through an equation like $F(x, y) = 0$ rather than an explicit formula $y = f(x)$. Many equations in multivariable calculus can't be neatly solved for one variable, so you differentiate the equation as it stands.

The core procedure for finding $\frac{dy}{dx}$ from $F(x, y) = 0$:

1. Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.
2. Every time you differentiate a term involving $y$, apply the chain rule and multiply by $\frac{dy}{dx}$.
3. Collect all terms containing $\frac{dy}{dx}$ on one side.
4. Solve for $\frac{dy}{dx}$.

This gives the shortcut formula directly:

$\frac{dy}{dx} = -\frac{F_x}{F_y}, \quad \text{provided } F_y \neq 0$

where $F_x = \frac{\partial F}{\partial x}$ and $F_y = \frac{\partial F}{\partial y}$. The negative sign is not optional; it comes from differentiating $F(x, y) = 0$ and isolating $\frac{dy}{dx}$.

Example: For $F(x,y) = x^2 + y^3 - 6xy = 0$, we get $F_x = 2x - 6y$ and $F_y = 3y^2 - 6x$, so

$\frac{dy}{dx} = -\frac{2x - 6y}{3y^2 - 6x}$

This extends naturally to three or more variables. If $F(x, y, z) = 0$ defines $z$ implicitly as a function of $x$ and $y$, then:

$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}$

again requiring $F_z \neq 0$.

Applying the Chain Rule and Total Differential

The formula above comes from the chain rule for partial derivatives. If $z = f(x, y)$ where $x = g(t)$ and $y = h(t)$, the chain rule gives:

$\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$

The total differential of $f(x, y)$ captures the same idea in differential form:

$df = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy$

This represents the infinitesimal change in $f$ resulting from infinitesimal changes in $x$ and $y$. For implicit differentiation, you set $dF = 0$ (since $F$ is constant along the curve), which immediately yields:

$\frac{\partial F}{\partial x}\,dx + \frac{\partial F}{\partial y}\,dy = 0 \quad \Longrightarrow \quad \frac{dy}{dx} = -\frac{F_x}{F_y}$

So the total differential is really the engine behind the shortcut formula.

Implicit Functions and Curves

Implicit Function Theorem and Level Curves

The Implicit Function Theorem makes the whole approach rigorous. It states: if $F(x, y) = 0$ at a point $(a, b)$, and if $F$ has continuous partial derivatives near $(a, b)$ with $\frac{\partial F}{\partial y}(a, b) \neq 0$, then there exists a unique differentiable function $y = f(x)$ defined near $x = a$ satisfying $F(x, f(x)) = 0$.

The condition $F_y \neq 0$ is critical. It's exactly the condition that appears in the denominator of $\frac{dy}{dx} = -\frac{F_x}{F_y}$. Where $F_y = 0$, the curve may have a vertical tangent or a self-intersection, and you can't locally express $y$ as a function of $x$.

Level curves connect directly to this framework. A level curve of $f(x, y)$ is the set of points where $f(x, y) = c$ for some constant $c$. Setting $F(x, y) = f(x, y) - c$, you're back to the $F = 0$ setup. Contour lines on a topographic map and isobars on a weather map are familiar examples of level curves.

Gradient Vector and Its Applications

The gradient vector of $f(x, y)$ is:

$\nabla f = \left(\frac{\partial f}{\partial x},\; \frac{\partial f}{\partial y}\right)$

Two key properties tie the gradient to implicit differentiation:

• $\nabla f$ points in the direction of greatest rate of increase of $f$ at a given point.
• $\nabla f$ is perpendicular (normal) to the level curve $f(x, y) = c$ at every point on that curve.

The perpendicularity property is why implicit differentiation works geometrically. Along a level curve, $f$ doesn't change, so any tangent vector to the curve must be orthogonal to $\nabla f$. If the level curve has tangent direction $(1, \frac{dy}{dx})$, then dotting with $\nabla f$ and setting the result to zero gives:

$f_x \cdot 1 + f_y \cdot \frac{dy}{dx} = 0 \quad \Longrightarrow \quad \frac{dy}{dx} = -\frac{f_x}{f_y}$

This is the same formula derived earlier, now with a geometric interpretation. The gradient is also central to optimization and to studying flows in physics, but for this section, its role as the normal to level curves is the main takeaway.