Derivatives of implicit functions

Some curves refuse to hand you a tidy formula "y=y= something in xx." A circle x2+y2=25x^2+y^2=25, or the tangled loop x3+y3=6xyx^3+y^3=6xy, ties xx and yy together in a single equation with no clean way to isolate yy. Yet you can still find the slope of the tangent at any point on such a curve — you simply differentiate the whole equation, term by term, treating yy as an unnamed function of xx throughout. Every time a yy-term is differentiated, the chain rule quietly attaches a factor of dydx\dfrac{dy}{dx}. Collect those terms, solve algebraically, and you have the slope — without ever solving for yy explicitly. This article builds that machinery from the chain rule you already know, checks it against a circle's known geometry, and extends it to a genuine two-variable curve.

The core insight

Differentiating both sides of an implicit equation term by term, and attaching a factor of dydx\dfrac{dy}{dx} to every yy-term via the chain rule, lets you solve algebraically for the slope of a curve you never had to rewrite as y=f(x)y=f(x).

The core idea

Why yy-terms pick up an extra factor. On a small enough piece of a curve like x2+y2=25x^2+y^2=25 (away from where it doubles back), yy behaves like some differentiable function of xx, even though we have no formula for it. Any term containing yy is therefore a composite expression — "an operation applied to the function yy of xx" — and differentiating a composite always calls the chain rule. That is the entire trick behind implicit differentiation: nothing new is invented, the ordinary chain rule is just applied to an unnamed function.

The two patterns you will use constantly. A pure power of yy differentiates like any power rule, but with the extra factor attached:

ddx(yn)=n yn−1dydx,ddx(sin⁡y)=cos⁡y dydx\frac{d}{dx}\big(y^n\big) = n\,y^{n-1}\frac{dy}{dx}, \qquad \frac{d}{dx}(\sin y) = \cos y\,\frac{dy}{dx}

where: nn is a fixed real exponent and dydx\dfrac{dy}{dx} is exactly the unknown you are trying to find — it appears on the right because yy is being treated as a function of xx, not a constant. A mixed term like xyxy or x2y3x^2y^3 needs the product rule combined with this chain-rule adjustment, since it is a product of an xx-part and a yy-part:

ddx(xy)=x⋅dydx+y⋅1\frac{d}{dx}(xy) = x\cdot\frac{dy}{dx} + y\cdot 1

where: the first piece comes from differentiating yy while xx is held as a factor (chain rule), and the second comes from differentiating xx while yy is held as a factor — the usual product-rule split.

The four-step procedure. (1) Differentiate both sides of the equation with respect to xx, term by term. (2) Apply the chain rule to every yy-term, and the product rule wherever xx and yy multiply. (3) Move every term containing dydx\dfrac{dy}{dx} to one side of the equation and everything else to the other side. (4) Factor out dydx\dfrac{dy}{dx} and divide to isolate it. The result will generally involve both xx and yy — that is normal, not a sign of an error, because a single implicit relation can pass through many points with many different slopes. The figure below shows why the circle example is such a good sanity check: the tangent line implicit differentiation produces at a point is always perpendicular to the radius drawn to that point, exactly as circle geometry from earlier classes predicts.

Figure
A circle with a radius drawn to a point on it and the tangent line at that point meeting the radius at a right angle
Implicit slope on a circle matches geometry: tangent is perpendicular to radius origin point (x0, y0) radius, slope y0/x0 tangent, slope minus x0/y0 Product of the two slopes is minus one, so the angle between them is a right angle
Derivation ATangent to a circle is perpendicular to the radius

The goal is to derive the slope of the tangent line to the circle x2+y2=r2x^2+y^2=r^2 at an arbitrary point (x0,y0)(x_0,y_0) on the circle using implicit differentiation, and then to show algebraically that this tangent slope, multiplied by the slope of the radius drawn to that same point, always equals −1-1 — confirming with pure algebra a geometric fact about circles you already accept on sight.

Step 1. Differentiate x2+y2=r2x^2+y^2=r^2 implicitly with respect to xx, remembering rr is a fixed constant and yy is a function of xx: 2x+2ydydx=02x + 2y\dfrac{dy}{dx} = 0.

Step 2. Solve for dydx\dfrac{dy}{dx}: dydx=−xy\dfrac{dy}{dx} = -\dfrac{x}{y}. This holds at every point on the circle where y≠0y\neq 0. Evaluated at the specific point (x0,y0)(x_0,y_0), the tangent slope is mtangent=−x0y0m_{\text{tangent}} = -\dfrac{x_0}{y_0}.

Step 3. Compute the slope of the radius joining the origin (0,0)(0,0) to (x0,y0)(x_0,y_0) using the ordinary two-point slope formula: mradius=y0−0x0−0=y0x0m_{\text{radius}} = \dfrac{y_0-0}{x_0-0} = \dfrac{y_0}{x_0}.

Step 4. Multiply the two slopes: mtangent⋅mradius=(−x0y0)(y0x0)=−1m_{\text{tangent}}\cdot m_{\text{radius}} = \left(-\dfrac{x_0}{y_0}\right)\left(\dfrac{y_0}{x_0}\right) = -1. Two lines whose slopes multiply to −1-1 are perpendicular, so the tangent to a circle at any point is perpendicular to the radius at that point — a result implicit differentiation reproduces purely algebraically, without ever drawing a picture. Bonus: the tangent line itself is y−y0=−x0y0(x−x0)y-y_0 = -\dfrac{x_0}{y_0}(x-x_0), which simplifies to xx0+yy0=r2xx_0+yy_0=r^2, the standard tangent-to-circle formula.

Worked example

Find the equation of the tangent line to the folium x3+y3=6xyx^3+y^3=6xy at the point (3,3)(3,3).

Step 1. Confirm (3,3)(3,3) actually lies on the curve: x3+y3=27+27=54x^3+y^3 = 27+27 = 54 and 6xy=6(3)(3)=546xy = 6(3)(3) = 54. Both sides match, so the point is valid.

Step 2. Differentiate the whole equation implicitly, exactly as in the core idea: ddx(x3)+ddx(y3)=ddx(6xy)\dfrac{d}{dx}(x^3)+\dfrac{d}{dx}(y^3) = \dfrac{d}{dx}(6xy) gives 3x2+3y2dydx=6(xdydx+y)=6xdydx+6y3x^2 + 3y^2\dfrac{dy}{dx} = 6\left(x\dfrac{dy}{dx}+y\right) = 6x\dfrac{dy}{dx}+6y.

Step 3. Collect the dydx\dfrac{dy}{dx} terms on one side: 3y2dydx−6xdydx=6y−3x23y^2\dfrac{dy}{dx} - 6x\dfrac{dy}{dx} = 6y-3x^2, so dydx(3y2−6x)=6y−3x2\dfrac{dy}{dx}(3y^2-6x) = 6y-3x^2, giving dydx=6y−3x23y2−6x=2y−x2y2−2x\dfrac{dy}{dx} = \dfrac{6y-3x^2}{3y^2-6x} = \dfrac{2y-x^2}{y^2-2x}.

Step 4. Evaluate the slope at (3,3)(3,3): numerator 2(3)−32=6−9=−32(3)-3^2 = 6-9=-3; denominator 32−2(3)=9−6=33^2-2(3) = 9-6=3. So dydx∣(3,3)=−33=−1\dfrac{dy}{dx}\Big|_{(3,3)} = \dfrac{-3}{3} = -1.

Step 5. Write the tangent line through (3,3)(3,3) with slope −1-1: y−3=−1(x−3)y-3 = -1(x-3), which simplifies to x+y=6x+y=6.

Competitive stretch

Notice the denominator y2−2xy^2-2x vanishes when y2=2xy^2=2x; substituting into the original curve equation shows this happens at (0,0)(0,0) and at one other point on the loop — exactly where the tangent line to the folium becomes vertical, since dydx\dfrac{dy}{dx} is undefined there. Spotting where an implicit slope formula's denominator vanishes is a quick way to locate vertical tangents without extra work.

Common traps

  • Forgetting the dydx\dfrac{dy}{dx} on a yy-term. Writing ddx(y2)=2y\dfrac{d}{dx}(y^2)=2y instead of 2ydydx2y\dfrac{dy}{dx} treats yy as a constant — the single most common slip.
  • Splitting a mixed term wrong. For xyxy or x2y3x^2y^3, both the product rule and the chain rule are needed; dropping either gives a wrong derivative.
  • Isolating dydx\dfrac{dy}{dx} too early. Try to solve before every term containing it has been moved to one side, and the algebra falls apart.
  • Rejecting an answer because it has both xx and yy in it. That is expected — an implicit derivative is generally a function of both variables, not a flaw to fix.

Ask the tutor: if collecting the dydx\dfrac{dy}{dx} terms and solving the algebra feels shaky, ask the AI tutor to walk through another implicit differentiation example with you line by line.

Why does every y-term get an extra dy by dx but every x-term does not?

Because xx is the variable we are differentiating with respect to, so ddx(xn)=nxn−1\dfrac{d}{dx}(x^n)=nx^{n-1} needs nothing extra — you are differentiating with respect to itself. But yy is being treated as an unnamed function of xx, so any term in yy is really a composite expression, and the chain rule always contributes a derivative of the "inner function," which here is dydx\dfrac{dy}{dx}. The rule is not arbitrary; it is the same chain rule you already use for something like sin⁡(3x)\sin(3x), just with y(x)y(x) standing in for the inner function instead of 3x3x.

Can I just solve for y first and use the ordinary derivative instead of doing this implicitly?

Sometimes, but it is often harder or impossible. For x2+y2=25x^2+y^2=25 you could solve y=±25−x2y=\pm\sqrt{25-x^2} and differentiate that, but you would need two cases for the two branches, and the algebra is messier than implicit differentiation. For x3+y3=6xyx^3+y^3=6xy there is no elementary way to isolate yy at all. Implicit differentiation sidesteps both problems by working directly with the equation as given, which is exactly why it is the standard tool for such curves.

My final slope has both x and y in it — is that wrong?

No, that is the normal and expected outcome. A single implicit equation like x3+y3=6xyx^3+y^3=6xy passes through many points, and each point generally has a different tangent slope, so the slope formula must depend on both coordinates. To get an actual number, always substitute the specific point you care about, as in the worked example above — never leave the answer as an unevaluated expression when a specific point was given.

Before this

Where this leads