AP Calculus AB and BC

Implicit Differentiation vs Related Rates

Q: Should I plug in the given values before differentiating?

No, substitute only after differentiating. A value such as x = 3 is true at one instant rather than for all time, so putting it in first turns a changing quantity into a constant and deletes the \frac{dx}{dt} term you needed.

Related rates is implicit differentiation performed with respect to time. Differentiating with respect to x gives dy/dx and leaves the x terms alone; differentiating with respect to t gives both dx/dt and dy/dt, so every changing letter picks up a chain factor instead of just one.

Implicit differentiation

Use when: You are handed an equation in x and y, nothing is moving, and the question asks for a slope, a tangent line, or a second derivative.

Related rates

Use when: The quantities change as time passes, the problem hands you one rate and asks for another, so time is the variable underneath everything.

Side by side

	Implicit differentiation	Related rates
Differentiate with respect to	$x$	$t$
Every $y$ term picks up	$\frac{dy}{dx}$	$\frac{dy}{dt}$
Every $x$ term picks up	$\frac{dx}{dx} = 1$ , so nothing appears on the page	$\frac{dx}{dt}$ , which has to be written every time
Where the equation comes from	Given to you as a curve	Built from the geometry: area, volume, Pythagoras, similar triangles
What the answer looks like	A slope, expressed in terms of $x$ and $y$	A number carrying units per unit of time

Take one equation and differentiate it two ways. The circle $x^2 + y^2 = 25$ describes a fixed curve, and differentiating with respect to $x$ treats $y$ as a function of $x$ and produces a slope. Now let a point travel along that circle so both coordinates are functions of time; differentiating with respect to $t$ produces a relationship between the two rates. Same equation, same chain rule, different variable underneath.

\text{From } x^2 + y^2 = 25: \qquad 2x + 2y\frac{dy}{dx} = 0 \qquad \text{versus} \qquad 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

The second version is the ladder problem. A $5$ metre ladder leans against a wall with its foot $x$ metres from the base and its top $y$ metres up, so $x^2 + y^2 = 25$ holds at every instant. If the foot slides out at $\frac{dx}{dt} = 2$ metres per second at the moment when $x = 3$ , then $y = 4$ , and substituting gives $2(3)(2) + 2(4)\frac{dy}{dt} = 0$ , so $\frac{dy}{dt} = -\frac{3}{2}$ metres per second. The negative sign reports that the top is sliding down.

The factor that goes missing

In implicit differentiation the $x$ terms carry $\frac{dx}{dx} = 1$ , so no visible factor appears and students learn to expect a chain factor on $y$ alone. Carried into related rates, that habit drops the $\frac{dx}{dt}$ and produces $2x + 2y\frac{dy}{dt} = 0$ , which returns $-\frac{3}{4}$ for the ladder instead of $-\frac{3}{2}$ . The rate you were given never entered the calculation.

Frequently asked questions

Is related rates just implicit differentiation?

Yes, with time as the variable you differentiate against. The chain rule, the product rule, and the habit of treating each letter as a function all carry over unchanged. The genuinely new work is building the equation from the geometry and attaching units to the answer.

Do I put a rate on every variable or only one?

Every letter that changes with time gets its own rate factor, so a product like $A = xy$ differentiates to $\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}$ . Quantities fixed for the whole problem, such as the length of the ladder, are constants and contribute nothing.

Should I plug in the given values before differentiating?

No, substitute only after differentiating. A value such as $x = 3$ is true at one instant rather than for all time, so putting it in first turns a changing quantity into a constant and deletes the $\frac{dx}{dt}$ term you needed.

In the CED: Unit 3: Chain Rule, Implicit, and Inverses, Unit 4: Contextual Applications