Module 4 · Lesson 4.4

Implicit Differentiation & Related Rates

Differentiate equations that link two changing quantities, then solve for the unknown rate when they move in lockstep.

Intuition

Sometimes a relationship is given by an equation that mixes $x$ and $y$ together, not by an explicit formula $y=f(x)$. Both variables can still change with time. Differentiating both sides with respect to $t$ produces an equation that links their rates $\frac{dx}{dt}$ and $\frac{dy}{dt}$. That is related-rates reasoning.

The classic picture is a ladder sliding down a wall: the top drops while the base slides out. Pythagoras keeps the length fixed, so the two speeds are locked together. Knowing one speed instantly gives the other.

Core concept

Implicit differentiation treats every variable as a function of time, applies the chain rule to each term, and solves the resulting equation for the unknown rate. The geometric constraint (fixed ladder length, expanding circle, \ldots) supplies the equation that binds the rates.

Ladderlength $L=5$ m
Constraint$x^{2}+y^{2}=25$
Differentiate$2x\,x'+2y\,y'=0$
At $x=3$, $y=4$$x'=0.6$ m/s out
Solve$y'=-0.45$ m/s

Interactive Lab

Slide the base distance $x$. The ladder stays length 5; the top height $y$ and both rates update from the differentiated constraint.

$x^{2}+y^{2}=25$ → $x\,x'+y\,y'=0$
Current rates will appear here.

Green point is the top of the ladder; amber is the base. The differentiated Pythagorean relation forces the two speeds to stay in lockstep.

Key Idea

$$ \frac{d}{dt}\bigl(F(x(t),y(t))\bigr)=0 \quad\Rightarrow\quad F_x\,x'+F_y\,y'=0 $$

When a constraint $F(x,y)=C$ holds for all time, its total derivative with respect to $t$ is zero. Solving for the unknown rate (usually $y'$) yields the related-rate formula. The same technique works for any implicit relation—circles, similar triangles, volume formulas—once every variable is viewed as a function of time.

$x(t),y(t)$Quantities that change with time
$x',y'$Their instantaneous rates $\frac{dx}{dt}$, $\frac{dy}{dt}$
$F(x,y)=C$The geometric or physical constraint that never changes
$F_x,F_y$Partial derivatives of the constraint (chain-rule coefficients)

Think Further

  1. Why does the top of the ladder speed up as it nears the ground, even if the base moves at constant speed?
  2. If a circular oil slick expands so that its radius grows at 2 cm/s, how would you find the rate at which the area is increasing when the radius is 30 cm?
  3. What goes wrong if you try to solve for $y'$ at the exact moment the ladder is horizontal?