AP Calculus AB and BC

Chain Rule vs Implicit Differentiation

Q: Is implicit differentiation a different rule from the chain rule?

No. It is the chain rule used on y, whose derivative is not known yet, so the inner factor stays written as \frac{dy}{dx}.

Q: Why do only the y terms pick up a dy/dx?

Because \frac{dx}{dx} = 1. Every term gets multiplied by the derivative of its inside function, and for a term in x that factor is 1, so it never shows.

Q: Can I use implicit differentiation on an ordinary function?

Yes, and it agrees. Differentiating y = x^2 implicitly gives \frac{dy}{dx} = 2x, which is what the power rule gives directly.

Implicit differentiation is not a separate rule; it is the chain rule applied to y, treated as an unknown function of x. The derivative of y that appears whenever you differentiate a term containing y is exactly the chain rule factor. Reach for it when the equation mixes x and y instead of naming the inside function.

Chain rule

Use when: The inside function is written out, as in $\sin(x^2)$ or $(3x + 1)^5$ , so its derivative can be computed on the spot.

Implicit differentiation

Use when: The inside function is $y$ itself, as in $x^2 + y^2 = 25$ , so its derivative has to stay written as $\frac{dy}{dx}$ .

Side by side

	Chain rule	Implicit differentiation
What it handles	A known function inside another	An unknown function of $x$ inside another
The inner derivative	Computable, such as $2x$ for $x^2$	Unknown, so it stays as $\frac{dy}{dx}$
Worked example	$\frac{d}{dx}\left[(f(x))^2\right] = 2f(x) \cdot f'(x)$	$\frac{d}{dx}\left[y^2\right] = 2y \cdot \frac{dy}{dx}$
How it finishes	The derivative is already isolated	Collect the $\frac{dy}{dx}$ terms, factor, then divide
Answer is in terms of	$x$ alone	Usually both $x$ and $y$
Common trap	Forgetting the inner derivative	Differentiating $y^2$ as $2y$

Write $y = f(x)$ and the two methods collapse into one. Differentiating $y^2$ means differentiating $(f(x))^2$ , which the chain rule turns into $2f(x) \cdot f'(x)$ . Renaming $f(x)$ back to $y$ and $f'(x)$ back to $\frac{dy}{dx}$ gives the implicit result, so nothing new was used.

\frac{d}{dx}\left[y^3\right] = 3y^2 \cdot \frac{dy}{dx}

The real difference shows up after the differentiating is done. An ordinary chain rule problem ends with the answer written out. An implicit problem ends with an equation in which $\frac{dy}{dx}$ is the unknown, so there is an algebra stage: gather those terms on one side, factor $\frac{dy}{dx}$ out, and divide.

Where the factor goes missing

Terms in $x$ alone never pick up a $\frac{dy}{dx}$ , and terms containing $y$ always do. Differentiating $x^2 + y^2 = 25$ gives $2x + 2y \cdot \frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = -\frac{x}{y}$ . Skip the factor and there is nothing left to solve for, which is the signal that a step was dropped.

Frequently asked questions

Is implicit differentiation a different rule from the chain rule?

No. It is the chain rule used on $y$ , whose derivative is not known yet, so the inner factor stays written as $\frac{dy}{dx}$ .

Why do only the y terms pick up a dy/dx?

Because $\frac{dx}{dx} = 1$ . Every term gets multiplied by the derivative of its inside function, and for a term in $x$ that factor is $1$ , so it never shows.

Can I use implicit differentiation on an ordinary function?

Yes, and it agrees. Differentiating $y = x^2$ implicitly gives $\frac{dy}{dx} = 2x$ , which is what the power rule gives directly.

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