AP Calculus AB and BC

Implicit Differentiation: Finding dy/dx Step by Step

Q: When should I use implicit differentiation instead of solving for y?

Use it whenever solving for y is impossible or messy: y appears to a power, sits inside a product with x, or is buried in a trig or exponential expression. If you can write y = f(x) cleanly, the ordinary rules are faster. If you cannot, differentiate the equation as it stands.

Q: Why does dy/dx appear when I differentiate a y-term?

Because y is a function of x. Differentiating y^2 is a chain-rule problem: the outer derivative 2y times the derivative of the inside, which is \frac{dy}{dx}. That \frac{dy}{dx} is the derivative of the inner function, and it never disappears.

Implicit differentiation finds dy/dx when an equation mixes x and y and you can't solve for y cleanly. Differentiate both sides with respect to x; each y-term picks up a dy/dx from the chain rule. Then gather the dy/dx terms, factor, and divide. Reach for it when y is trapped, like x^2 + y^2 = 25.

Spotting when to use implicit differentiation

Implicit differentiation is the tool when an equation ties $x$ and $y$ together and you cannot (or would rather not) solve for $y$ first. A circle such as $x^2 + y^2 = 25$ is the classic signal: $y$ appears squared, so isolating it forces a $\pm$ and breaks the curve into two separate halves. Equations like $x^3 + y^3 = 6xy$ or $\sin(xy) = x$ are worse still, because $y$ is tangled past anything clean algebra can undo. These are relations, not functions, and many of them fail the vertical line test outright.

The recognition question is short. Can you rewrite the equation as $y = f(x)$ without a fight? If yes, do exactly that and differentiate with the ordinary rules from Units 2 and 3. If $y$ is trapped, mixed into products with $x$ , or raised to powers you cannot unwind, switch to implicit differentiation. You never isolate $y$ at all. You differentiate the equation as written and solve for the slope $\frac{dy}{dx}$ directly.

Equation	Can you isolate $y$ ?	Best method
$y = x^2\sin x$	Already solved	Ordinary rules (product rule)
$x^2 + y^2 = 25$	Only with a $\pm$	Implicit differentiation
$x^3 + y^3 = 6xy$	No clean way	Implicit differentiation
$xy = 4$	Yes, $y = \frac{4}{x}$	Either works

The one idea behind the method

Every $y$ in the equation stands for an unnamed function of $x$ . So every $y$ you differentiate carries a hidden $\frac{dy}{dx}$ , supplied by the chain rule. That is the entire basis of Topic 3.2, and it is why the chain rule from Topic 3.1 has to come first.

The method: differentiate, then solve for dy/dx

Differentiate both sides of the equation with respect to $x$ . Terms built only from $x$ follow the rules you already know, and a constant differentiates to $0$ . The moment you differentiate a term containing $y$ , the chain rule fires. The derivative of $y^2$ is not $2y$ ; it is $2y\frac{dy}{dx}$ , because you differentiate the outer power and then multiply by the derivative of the inside function: the inside is $y$ , and its derivative is $\frac{dy}{dx}$ . The general pattern is worth memorizing:

\frac{d}{dx}\left(y^n\right) = n y^{n-1}\frac{dy}{dx}

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .
Apply the chain rule to every $y$ -term, attaching a factor of $\frac{dy}{dx}$ ; use the product rule wherever $x$ and $y$ are multiplied.
Move every term containing $\frac{dy}{dx}$ to one side and everything else to the other.
Factor out $\frac{dy}{dx}$ and divide to isolate it.

Step four explains why the result almost always contains both $x$ and $y$ . That is not a mistake to fix. The slope of an implicit curve genuinely depends on which point you sit at, and naming a point on such a curve takes both coordinates. An answer like $\frac{dy}{dx} = -\frac{x}{y}$ is a finished answer.

When a problem asks for the slope at a specific point, wait until after you have solved for $\frac{dy}{dx}$ , then substitute both coordinates. Substituting too early, before the derivative is isolated, throws away the $\frac{dy}{dx}$ you are trying to find.

The mistakes that cost points

The single most common error is differentiating a $y$ -term as if $y$ were $x$ . Writing $\frac{d}{dx}(y^3) = 3y^2$ silently drops the $\frac{dy}{dx}$ and corrupts every step after it. Every $y$ you touch earns its chain-rule factor, without exception.

The second trap shows up whenever $x$ and $y$ are multiplied, like the $xy$ in $x^2 + xy + y^2 = 7$ . That term needs the product rule, and the $y$ -half of the product still needs its own $\frac{dy}{dx}$ . The correct expansion is $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ . Skipping the product rule here is one of the most frequent free-response deductions in Unit 3.

The third slip is algebraic. After you collect the $\frac{dy}{dx}$ terms, factor $\frac{dy}{dx}$ out before you divide. Trying to divide term by term instead of factoring first is where a correct derivative turns into a wrong final fraction.

Extending to the second derivative

To find $\frac{d^2y}{dx^2}$ , differentiate the first derivative again with respect to $x$ . Two details do the damage. First, the first derivative usually still contains $y$ , so differentiating it calls for the chain rule (and often the quotient rule) a second time. Second, any $\frac{dy}{dx}$ that reappears in this pass should be replaced with the expression you already found, after which you simplify, often using the original equation to clean things up. Topic 3.6 formalizes higher-order derivatives; the circle example below runs the whole process end to end.

This is also why a second-derivative answer for a circle can collapse to something as tidy as $-\frac{25}{y^3}$ : the original relation absorbs the $x^2 + y^2$ that the algebra produces.

Worked examples

Worked example

The circle: a clean first derivative

Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = 25$ , then find the slope at the point $(3, 4)$ .

Differentiate both sides with respect to $x$ , treating $y$ as an unknown function of $x$ : $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$ .
The $x^2$ term gives $2x$ by the power rule. The $y^2$ term needs the chain rule, so $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ . The constant $25$ differentiates to $0$ , leaving $2x + 2y\frac{dy}{dx} = 0$ .
Move the $x$ -term across and divide by $2y$ : $2y\frac{dy}{dx} = -2x$ , so $\frac{dy}{dx} = -\frac{x}{y}$ .
Now substitute the point $(3, 4)$ , after the derivative is isolated: $\frac{dy}{dx} = -\frac{3}{4}$ .

$\frac{dy}{dx} = -\frac{x}{y}$ , which is $-\frac{3}{4}$ at $(3, 4)$ .

Worked example

A product term: xy needs the product rule

Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$ .

Differentiate each term with respect to $x$ . The middle term $xy$ is a product of two functions of $x$ , so flag it for the product rule: $\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0$ .
The power rule gives $\frac{d}{dx}(x^2) = 2x$ . The product rule gives $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ , since $\frac{d}{dx}(x) = 1$ . The chain rule gives $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ .
Assemble the pieces: $2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0$ .
Gather the $\frac{dy}{dx}$ terms on one side and move the rest to the other: $x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y$ .
Factor out $\frac{dy}{dx}$ , then divide: $\frac{dy}{dx}(x + 2y) = -(2x + y)$ , so $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$ .

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$ .

Worked example

The second derivative of the circle

Find $\frac{d^2y}{dx^2}$ for the circle $x^2 + y^2 = 25$ .

Start from the first derivative already found for this circle, $\frac{dy}{dx} = -\frac{x}{y}$ .
Differentiate again with respect to $x$ . The right side is a quotient, so use the quotient rule, remembering that $\frac{d}{dx}(y) = \frac{dy}{dx}$ : $\frac{d^2y}{dx^2} = -\frac{(1)(y) - x\frac{dy}{dx}}{y^2}$ .
Substitute the known $\frac{dy}{dx} = -\frac{x}{y}$ : $\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}$ .
Combine the numerator over $y$ : $y + \frac{x^2}{y} = \frac{y^2 + x^2}{y}$ , so $\frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^3}$ .
The original equation says $x^2 + y^2 = 25$ , so the result simplifies to $\frac{d^2y}{dx^2} = -\frac{25}{y^3}$ .

$\frac{d^2y}{dx^2} = -\frac{25}{y^3}$ .

Frequently asked questions

When should I use implicit differentiation instead of solving for y?

Use it whenever solving for $y$ is impossible or messy: $y$ appears to a power, sits inside a product with $x$ , or is buried in a trig or exponential expression. If you can write $y = f(x)$ cleanly, the ordinary rules are faster. If you cannot, differentiate the equation as it stands.

Why does dy/dx appear when I differentiate a y-term?

Because $y$ is a function of $x$ . Differentiating $y^2$ is a chain-rule problem: the outer derivative $2y$ times the derivative of the inside, which is $\frac{dy}{dx}$ . That $\frac{dy}{dx}$ is the derivative of the inner function, and it never disappears.

Why does my answer have both x and y in it?

That is normal and correct for implicit curves. The slope changes from point to point, and a point on a relation like $x^2 + y^2 = 25$ needs both an $x$ and a $y$ to pin it down. To get a number, substitute a specific point after solving for $\frac{dy}{dx}$ .