Implicit Differentiation Explained with Examples

What's the slope of a circle?

Not "at the top" or "at the side," those you could probably guess. What's the slope at some arbitrary point, using nothing but the equation $x^2 + y^2 = 25$?

You can't just differentiate this the way you differentiate $y = x^2$. There's no "y = ..." to start from. And yet the circle obviously has a slope everywhere except two points. Implicit differentiation is how we get at it anyway.

Why can't we just solve for y?

Some equations hand you $y$ directly. $y = 3x + 1$. $y = x^2 - 4$. These are called explicit functions, and differentiating them is routine.

But $x^2 + y^2 = 25$ doesn't do that. Solve for $y$ and you get $y = \pm\sqrt{25 - x^2}$, two separate formulas glued together at $x = \pm 5$. Worse, some equations can't be solved for $y$ at all, no matter how hard you push the algebra. We'll meet one of those later in this post.

So here's the question: can we find $\frac{dy}{dx}$ without ever isolating $y$?

Turns out, yes. And the trick is smaller than you'd expect.

The trick: differentiate both sides

Here's the whole idea in one sentence: differentiate both sides of the equation with respect to $x$, and every time you differentiate a term containing $y$, remember that $y$ is secretly a function of $x$.

That "remembering" is just the chain rule. Take $x^2 + y^2 = 25$. Differentiate term by term:

$\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]$

The first term is easy: $\frac{d}{dx}[x^2] = 2x$. The last term is zero, since 25 never changes. The middle term is where it gets interesting. $y^2$ isn't just "a square," it's the square of some function of $x$. The chain rule says differentiating it gives $2y$, times the rate $y$ itself is changing:

$2x + 2y\frac{dy}{dx} = 0$

Now it's just algebra. Isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{x}{y}$

That's it. A formula for the slope of a circle at any point $(x, y)$ on it, without ever writing $y$ as an explicit function of $x$. Try it below.

Watch what happens near the top and bottom of the circle. The formula $-x/y$ demands you divide by $y$, and right where $y = 0$, that blows up. Notice this is exactly where the explicit form would need to switch between $+\sqrt{25-x^2}$ and $-\sqrt{25-x^2}$. Same breakdown, seen from two angles.

And notice the tangent is always perpendicular to the radius, exactly as geometry says it should be. That's not a coincidence, it's a nice sanity check that the formula is telling the truth.

Where does that extra dy/dx come from?

Let's slow down on that one step, because it's the entire method.

Think of $y$ as a dial connected to $x$ by a hidden wire. You never see the wire's exact shape, but every time $x$ moves a little, $y$ moves too, by some amount that depends on the wire. That amount, per unit of $x$, is exactly $\frac{dy}{dx}$.

Now suppose a term in your equation is $y^2$. If $y$ were an independent variable, differentiating $y^2$ with respect to itself would just give $2y$. But $y$ isn't independent here, it's riding on $x$. So the chain rule inserts the wire's own rate: $\frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}$.

Every single time $y$ shows up in an equation, differentiating it costs you one factor of $\frac{dy}{dx}$. Plain $x$ terms never do, because $x$ is the thing we're differentiating with respect to. Its own rate of change with respect to itself is just $1$, so it never shows up explicitly.

What about terms where x and y multiply?

Circles are the easy case, because $x$ and $y$ never touch each other directly. But plenty of equations have a term like $xy$, where both variables are tangled together in the same product. That needs the product rule on top of the chain rule.

Take a simple one: $xy = 6$, a hyperbola. Differentiate both sides:

$\frac{d}{dx}[xy] = \frac{d}{dx}[6]$

The right side is $0$. The left side needs the product rule, treating $x$ and $y$ as two changing quantities multiplied together:

$x\frac{dy}{dx} + y \cdot 1 = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{y}{x}$

But why does a product of two changing things differentiate that way? Picture $xy$ as the area of a rectangle with width $x$ and height $y$. Nudge $x$ up by a small amount $\Delta x$, while staying on the curve so the area stays exactly $6$. The height $y$ has to shrink to compensate, by some $\Delta y$.

The new area comes from three pieces: a strip of width $\Delta x$ along the side (area $y \cdot \Delta x$), a strip of height $\Delta y$ along the top (area $x \cdot \Delta y$), and a tiny leftover corner (area $\Delta x \cdot \Delta y$). Since the total area never changes, those three pieces must sum to zero.

Now shrink $\Delta x$ toward zero. The two strips shrink proportionally, but the corner shrinks like $\Delta x \cdot \Delta y$, a product of two small things, so it vanishes much faster. What's left in the limit is exactly $y\,dx + x\,dy = 0$... which is the product rule, discovered from a rectangle instead of memorized from a table.

A curve you truly cannot solve for y

Here's the payoff. Consider:

$x^3 + y^3 = 6xy$

This is the Folium of Descartes, and unlike the circle, there is no clean way to write $y$ as a formula in $x$. It's a genuine loop, doubling back on itself, failing the vertical line test entirely. Explicit calculus has nothing to offer here. Implicit differentiation doesn't blink.

Differentiate both sides, using the chain rule on $y^3$ and the product rule on $6xy$:

$3x^2 + 3y^2\frac{dy}{dx} = 6\left(x\frac{dy}{dx} + y\right)$

Collect every $\frac{dy}{dx}$ term on one side:

$\frac{dy}{dx}\left(3y^2 - 6x\right) = 6y - 3x^2 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$

One formula, and it works at every point on that loop, explicit form or not.

Trace the loop slowly. Near the origin, the curve crosses itself, two different branches passing through the same point with two different tangent directions. The formula genuinely breaks down there ($0/0$), because at a self-crossing, "the" tangent line isn't a single thing anymore. Everywhere else, it just works.

What can go wrong

A few habits will trip you up, all fixable with practice.

Forgetting the $\frac{dy}{dx}$. If you differentiate $y^2$ and just write $2y$, you've silently pretended $y$ is a plain constant with respect to $x$. It isn't. Any term with $y$ in it needs that extra factor.

Skipping the product rule. A term like $xy$ or $3x^2y$ has both variables in it. Differentiate it as a product, not as if only one side were changing.

Losing track of $\frac{dy}{dx}$ during algebra. After differentiating, you'll usually have $\frac{dy}{dx}$ scattered across several terms. Collect them all onto one side before you factor and divide, or you'll drop one by accident.

None of these are new rules. They're the ordinary chain rule and product rule, applied carefully, term by term.

Try it yourself

Four different curves, one method. Pick one below, slide the point along it, and check that the formula and the picture always agree.

The short version

Implicit differentiation is what you get when you apply the ordinary chain rule to an equation instead of a lone function. Differentiate both sides with respect to $x$, treat every $y$ as a hidden function of $x$ (which costs a factor of $\frac{dy}{dx}$ each time it appears), use the product rule wherever $x$ and $y$ multiply together, then collect all the $\frac{dy}{dx}$ terms and solve for it algebraically. The result is a slope formula that works on circles, hyperbolas, loops, and any other curve defined by an equation, whether or not you could ever solve that equation for $y$ in the first place.