Implicit and parametric differentiation

Not every curve hands you a tidy $y=f(x)$ . A circle, an ellipse, or a tangled shape like $x^3+y^3=3xy$ mixes $x$ and $y$ together so thoroughly that solving for $y$ first is either impossible or a mess of square roots. Other curves — a projectile's path, a point tracing a circle, an astroid — are naturally described by giving $x$ and $y$ separately in terms of a third variable, a parameter, usually $t$ or $\theta$ . Implicit and parametric differentiation are the two techniques that let you find $\dfrac{dy}{dx}$ in both situations without ever forcing the curve into an explicit $y=f(x)$ form. Both rest on the same idea you already know: the chain rule, applied carefully to whichever variable is doing the "hiding."

The core insight

Differentiate implicit relations term-by-term with the chain rule (inserting $\dfrac{dy}{dx}$ wherever $y$ appears) and then solve algebraically for $\dfrac{dy}{dx}$ , or, for parametric curves, form $\dfrac{dy}{dx}$ directly as the ratio $\dfrac{dy/dt}{dx/dt}$ .

The core idea

Implicit differentiation applies when $y$ is not isolated on one side of the equation — for example $x^2+y^2=25$ or $x^3+y^3=3xy$ . Because $y$ is still secretly a function of $x$ , every term containing $y$ must be differentiated using the chain rule, treating $y$ as $y(x)$ :

\frac{d}{dx}\big[g(y)\big] = g'(y)\cdot\frac{dy}{dx}

where: $g(y)$ is any function built from $y$ (such as $y^3$ or $\sin y$ ), and the extra factor $\dfrac{dy}{dx}$ appears because $y$ itself depends on $x$ . Differentiate both sides of the given equation term by term, collect all the $\dfrac{dy}{dx}$ terms on one side, and solve.

Parametric differentiation applies when both $x$ and $y$ are given separately as functions of a third variable, typically $t$ : $x=x(t)$ , $y=y(t)$ . Then

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt}\ne0

where: $dy/dt$ and $dx/dt$ are the ordinary derivatives of $y$ and $x$ with respect to the parameter $t$ , computed separately and then divided. A curve like the folium of Descartes below cannot be written as one explicit $y=f(x)$ formula over its whole domain — implicit differentiation sidesteps that entirely by differentiating the equation exactly as given, term by term, and solving for $\dfrac{dy}{dx}$ only at the end. Parametric equations solve a related but different problem: they describe a curve traced out as $t$ (often time, or an angle $\theta$ ) increases, which is why the resulting slope often simplifies neatly using trigonometric identities.

Figure

The folium of Descartes curve with a tangent line drawn at the point one point five one point five, alongside a small astroid curve traced by an angle theta, showing one curve handled implicitly and the other parametrically

Common traps

Forgetting the $\dfrac{dy}{dx}$ factor when differentiating a $y$ -term implicitly — writing $\dfrac{d}{dx}(y^3)=3y^2$ instead of $3y^2\dfrac{dy}{dx}$ is the single most common slip.
Mishandling mixed terms like $xy$ or $x^2y$ — these need the product rule together with the chain rule on the $y$ part, not just one or the other.
In parametric problems, inverting the ratio — computing $\dfrac{dx/dt}{dy/dt}$ instead of $\dfrac{dy/dt}{dx/dt}$ .
Trying to rearrange an implicit relation into $y=f(x)$ before differentiating — often impossible or needlessly messy; differentiate the equation exactly as given.

Derivation AWhy dy/dx equals (dy/dt) over (dx/dt) for a parametric curve

The goal is to show, starting from nothing but the ordinary single-variable chain rule, that when a curve is given parametrically as $x=x(t)$ and $y=y(t)$ , the slope of the curve — the rate $y$ changes with $x$ — can be found without ever eliminating $t$ to write $y$ explicitly in terms of $x$ , simply by dividing the two separate derivatives $dy/dt$ and $dx/dt$ with respect to the parameter.

Step 1. Treat $y$ as a function of $x$ , and $x$ in turn as a function of $t$ , so that $y$ is really a composite function of $t$ through $x$ : $y = y\big(x(t)\big)$ . This composite structure is exactly the setup the chain rule is built for.

Step 2. Apply the chain rule to differentiate this composite with respect to $t$ . Since $y$ depends on $t$ only through $x$ , the chain rule gives

\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}.

This single line is the entire content of the derivation — everything else is rearrangement.

Step 3. Solve this equation for $\dfrac{dy}{dx}$ , the quantity we actually want, by dividing both sides by $\dfrac{dx}{dt}$ (valid whenever $\dfrac{dx}{dt}\ne0$ , i.e. the curve is not momentarily moving straight up or down as $t$ varies):

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

Step 4. Read the result. Both $dy/dt$ and $dx/dt$ are ordinary derivatives of single-variable functions — no special new rule is needed to compute them, only the standard differentiation techniques applied to $x(t)$ and $y(t)$ separately. Dividing the two gives the slope of the curve at each value of $t$ , directly, with no need to eliminate $t$ and produce an explicit $y=f(x)$ first.

Worked example

Find $\dfrac{dy}{dx}$ for the folium of Descartes $x^3+y^3=3xy$ , then use it to find the equation of the tangent line at the point $(1.5,1.5)$ , which lies on the curve since $1.5^3+1.5^3=6.75=3(1.5)(1.5)$ .

Step 1. Differentiate both sides with respect to $x$ , treating $y$ as $y(x)$ :

\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3xy)

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

using the chain rule on $y^3$ and the product rule on $3xy$ .

Step 2. Expand the right side and collect every $\dfrac{dy}{dx}$ term on one side:

3x^2 + 3y^2\frac{dy}{dx} = 3x\frac{dy}{dx} + 3y

\frac{dy}{dx}\big(3y^2-3x\big) = 3y-3x^2

Step 3. Solve for $\dfrac{dy}{dx}$ and simplify by dividing through by $3$ :

\frac{dy}{dx} = \frac{y-x^2}{y^2-x}

Step 4. Substitute the point $(1.5,1.5)$ to get the slope there: $y-x^2 = 1.5-2.25=-0.75$ and $y^2-x = 2.25-1.5=0.75$ , so

\frac{dy}{dx}\Big|_{(1.5,1.5)} = \frac{-0.75}{0.75} = -1

Step 5. Write the tangent line using point-slope form, $y-y_1=m(x-x_1)$ with $m=-1$ , $(x_1,y_1)=(1.5,1.5)$ : $y-1.5=-1(x-1.5)$ , which simplifies to $x+y=3$ , the tangent line to the folium at $(1.5,1.5)$ .

Competitive stretch

For the astroid $x=a\cos^3\theta$ , $y=a\sin^3\theta$ , the parametric formula gives $\dfrac{dy}{dx}=-\tan\theta$ (dividing $dy/d\theta = 3a\sin^2\theta\cos\theta$ by $dx/d\theta=-3a\cos^2\theta\sin\theta$ and cancelling $3a\sin\theta\cos\theta$ ). At $\theta=\pi/4$ the point is $\left(\dfrac{a}{2\sqrt2},\dfrac{a}{2\sqrt2}\right)$ and the slope is $-\tan(\pi/4)=-1$ , giving the tangent line $x+y=\dfrac{a}{\sqrt2}$ — the same slope $-1$ as the folium example above, a useful pattern-check rather than a coincidence to rely on in general.

Common exam pitfalls (quick recap)

Leaving the final implicit answer with $y$ terms un-collected instead of solving algebraically for a single expression in $\dfrac{dy}{dx}$ .
Substituting the point into the original equation instead of into the derivative when asked for a slope or tangent line.

Ask the tutor: not sure whether a given relation is more efficient to handle implicitly or by first eliminating a parameter — bring your specific equation to the AI tutor and compare both approaches side by side.

Why can't I just solve x cubed plus y cubed equals 3xy for y and then differentiate normally?

You could try, but solving this relation explicitly for $y$ in terms of $x$ is either impossible in closed form or produces a horribly messy expression with cube roots, because $y$ appears in more than one term in a way ordinary algebra cannot cleanly untangle. Implicit differentiation avoids this completely: you never solve for $y$ first. You differentiate the equation exactly as it stands, term by term, using the chain rule wherever $y$ appears, and only solve algebraically at the very last step — for $\dfrac{dy}{dx}$ itself, not for $y$ .

In the parametric formula, why does dividing dy by dt by dx by dt actually give the slope dy by dx?

Because $y$ , seen as a function of $x$ , is secretly a composite function of the parameter $t$ through $x$ : $y=y(x(t))$ . The chain rule for this composite says $dy/dt = (dy/dx)\cdot(dx/dt)$ . Rearranging that single equation — dividing both sides by $dx/dt$ — gives $dy/dx = (dy/dt)/(dx/dt)$ directly. So the parametric formula is not a new rule at all; it is the ordinary chain rule, just solved for the piece we want.

Do I always need to find dy by dx in general form, or can I sometimes just plug in a value of t or theta straight away?

If a question only asks for the slope or tangent at one specific point, it is often faster to differentiate $x(t)$ and $y(t)$ , substitute the given value of $t$ (or $\theta$ ) into each derivative separately, and then divide the two numbers — you don't have to simplify $dy/dx$ into a clean general formula first. However, if the question asks for the slope "in terms of $t$ " or wants a general expression (as in the astroid example, where $-\tan\theta$ is a genuinely useful simplified form), you do need the full simplified ratio, and trigonometric identities are usually the key to that simplification.