Derivatives of a Function in Parametric Form Explained

What Is The Derivative Of A Function In Parametric Form?

The derivative of a function in parametric form is the slope $\frac{dy}{dx}$ of a curve whose points are given by two equations, $x = x(t)$ and $y = y(t)$, in terms of a third variable $t$ called the parameter. You find it by differentiating both equations with respect to $t$ and dividing:

$$\frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}, \qquad \frac{dx}{dt} \neq 0$$

Here $\frac{dy}{dt}$ is how fast $y$ changes as $t$ grows, $\frac{dx}{dt}$ is how fast $x$ changes, and their ratio is the change in $y$ per unit change in $x$. The formula needs two conditions: both $x(t)$ and $y(t)$ must be differentiable at the value of $t$ you care about, and $\frac{dx}{dt}$ must not be zero there. When $\frac{dx}{dt} = 0$ the formula breaks, and that is exactly where vertical tangents and cusps live.

The equations themselves, how to sketch them and how to convert them, have their own page on parametric equations. This page is about the calculus that sits on top: the first derivative, the second derivative, and the tangents they produce. All derivatives here are written in Leibniz notation, so the variable each one is taken with respect to is always visible.

Why Does The Parametric Formula Work?

The parametric formula is the chain rule solved for the slope. Suppose that near some value of $t$, the curve can be read as $y$ depending on $x$, and $x$ depending on $t$. The chain rule then says:

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

Divide both sides by $\frac{dx}{dt}$, which is allowed only when it is not zero:

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

The condition $\frac{dx}{dt} \neq 0$ is doing real work here, not just avoiding division by zero. If $\frac{dx}{dt}$ is not zero, then $x$ is strictly increasing or strictly decreasing in $t$ near that point, so each $x$ comes from exactly one $t$ and $y$ really is a function of $x$ there. That local reversal of $t$ and $x$ is the same idea behind inverse functions and differentiation, where $\frac{dt}{dx} = \frac{1}{dx/dt}$.

So the full recipe is three lines:

1. Differentiate $x(t)$ to get $\frac{dx}{dt}$.

2. Differentiate $y(t)$ to get $\frac{dy}{dt}$.

3. Divide the second by the first, and simplify.

What Does The Parametric Derivative Mean On A Curve?

On a parametric curve, $\frac{dy}{dx}$ is the slope of the tangent line, while the pair $\left(\frac{dx}{dt}, \frac{dy}{dt}\right)$ is the velocity of the moving point. A question on Quora asks exactly this: what is the interpretation of $\frac{dy}{dx}$ in parametric equations, and why is it different from the velocity?

The answer is that the velocity is an arrow and the slope is only its direction. Picture $t$ as time and the point $(x(t), y(t))$ as a car on a road:

• The velocity vector $\left(\frac{dx}{dt}, \frac{dy}{dt}\right)$ points along the road and its length is the car's speed.

• The slope $\frac{dy}{dx}$ is the rise over run of that arrow. It tells you which way the road tilts, but not how fast the car is going.

• Same road, different speed: trace the curve twice as fast (replace $t$ by $2t$) and both components of velocity double, but their ratio, the slope, stays the same.

That last point is a useful check on any answer. The slope belongs to the curve, so it cannot depend on how quickly you travel along it. The full motion story, with speed and acceleration, is on the velocity and acceleration page.

Why Not Just Eliminate The Parameter?

Eliminating the parameter means solving one equation for $t$ and substituting into the other, so the curve becomes an ordinary $y = f(x)$. Brainly carries the textbook version of the instruction: "if x and y are connected parametrically by the given equations, then without eliminating the parameter find $\frac{dy}{dx}$". There are three good reasons the parametric formula is usually the better route.

• Elimination is often impossible in closed form. For the cycloid, $x = r(\theta - \sin\theta)$ cannot be solved for $\theta$ with ordinary functions, so there is no tidy $y = f(x)$ to differentiate.

• Elimination can split one curve into pieces. The unit circle $x = \cos t$, $y = \sin t$ becomes $y = \pm\sqrt{1 - x^2}$, two functions with a sign choice to track. The parametric slope $-\frac{\cos t}{\sin t}$ handles the whole circle at once.

• The answer is wanted in terms of $t$ anyway. Tangent problems usually name a point by its parameter value, so a slope written in $t$ is the slope you need.

When elimination is easy, it makes a good check.

Take $x = \ln t$ and $y = t^2$ with $t > 0$. Parametrically, $\frac{dy}{dx} = \frac{2t}{1/t} = 2t^2$. Eliminating, $t = e^x$, so $y = e^{2x}$ and $\frac{dy}{dx} = 2e^{2x} = 2t^2$. Both routes agree, as they must.

How Do You Find The Second Derivative In Parametric Form?

The second derivative in parametric form is the derivative of $\frac{dy}{dx}$ with respect to $x$, and since $\frac{dy}{dx}$ is written in $t$, you differentiate it with respect to $t$ and divide by $\frac{dx}{dt}$ once more:

$$\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$

This is the same chain-rule move as before, applied to a new function. Call the first derivative $m(t) = \frac{dy}{dx}$. To find $\frac{dm}{dx}$, use $\frac{dm}{dt} = \frac{dm}{dx}\cdot\frac{dx}{dt}$ and divide. Expanding $\frac{d}{dt}$ of the quotient gives an equivalent one-line form:

$$\frac{d^2y}{dx^2} = \frac{\dfrac{dx}{dt},\dfrac{d^2y}{dt^2} - \dfrac{dy}{dt},\dfrac{d^2x}{dt^2}}{\left(\dfrac{dx}{dt}\right)^3}$$

The general ideas behind the second derivative carry over unchanged. A positive value means the curve is concave up at that point (bending upward as $x$ increases), and a negative value means concave down.

What it is not is $\frac{d^2y}{dt^2}$ divided by $\frac{d^2x}{dt^2}$. That ratio looks like the natural extension of the first-derivative rule, and it is wrong almost everywhere; Example 3 below lets it fail on a circle, where the right sign is obvious from the picture. Higher orders follow the same pattern: each new derivative is $\frac{d}{dt}$ of the previous one, divided by $\frac{dx}{dt}$.

How Do You Find Horizontal And Vertical Tangents?

Horizontal and vertical tangents come straight from the numerator and denominator of the slope formula. Because $\frac{dy}{dx}$ is a fraction, you read its zeros and its blow-ups from the top and the bottom separately:

The third row deserves a careful look. When both rates are zero, the moving point has stopped, and the shape of the curve there depends on how it starts moving again. You settle it by finding the limit of $\frac{dy}{dx}$ as $t \to t_0$; for the cycloid that limit is infinite, which gives the sharp, vertical-tipped cusp in the opening picture.

Once you have a slope $m$ at a parameter value $t_0$, the tangent line through the point $(x_0, y_0) = (x(t_0), y(t_0))$ is the usual point-slope line, covered in full on the tangent line equations page:

$$y - y_0 = m,(x - x_0)$$

Examples Of Derivatives Of A Function In Parametric Form

These six examples run from a one-line slope to a curve with two tangents at the same point.

Example 1

Find $\frac{dy}{dx}$ for $x = t^3 + 1$, $y = t^2 + 4t$, and evaluate it at $t = 1$.

$$\frac{dx}{dt} = 3t^2$$

$$\frac{dy}{dt} = 2t + 4$$

$$\frac{dy}{dx} = \frac{2t + 4}{3t^2} = \frac{2(t + 2)}{3t^2}$$

$$\left.\frac{dy}{dx}\right|_{t = 1} = \frac{2(3)}{3} = 2$$

Final answer: $\frac{dy}{dx} = \frac{2(t + 2)}{3t^2}$, which equals $2$ at $t = 1$ (the point $(2, 5)$). At $t = 0$ the formula is undefined, because $\frac{dx}{dt} = 0$ there.

Example 2

Find the tangent line to the ellipse $x = 3\cos t$, $y = 2\sin t$ at $t = \frac{\pi}{4}$.

The ellipse here has semi-axes 3 and 2. Differentiate each coordinate:

$$\frac{dx}{dt} = -3\sin t, \qquad \frac{dy}{dt} = 2\cos t$$

$$\frac{dy}{dx} = \frac{2\cos t}{-3\sin t} = -\frac{2}{3}\cot t$$

At $t = \frac{\pi}{4}$, $\cot\frac{\pi}{4} = 1$, so the slope is $-\frac{2}{3}$. The point is:

$$x_0 = 3\cos\frac{\pi}{4} = \frac{3\sqrt{2}}{2}, \qquad y_0 = 2\sin\frac{\pi}{4} = \sqrt{2}$$

$$y - \sqrt{2} = -\frac{2}{3}\left(x - \frac{3\sqrt{2}}{2}\right)$$

$$y = -\frac{2}{3}x + 2\sqrt{2}$$

Check with the standard ellipse tangent $\frac{x x_0}{9} + \frac{y y_0}{4} = 1$: it gives $2x + 3y = 6\sqrt{2}$, the same line.

Final answer: $y = -\frac{2}{3}x + 2\sqrt{2}$.

Example 3

For the unit circle $x = \cos(\pi t)$, $y = \sin(\pi t)$, find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$, then evaluate the second derivative at $t = \frac{1}{4}$.

This is a Brainly question almost word for word, and it is the cleanest place to watch the tempting shortcut fail.

The first derivative is routine:

$$\frac{dx}{dt} = -\pi\sin(\pi t), \qquad \frac{dy}{dt} = \pi\cos(\pi t)$$

$$\frac{dy}{dx} = \frac{\pi\cos(\pi t)}{-\pi\sin(\pi t)} = -\cot(\pi t)$$

The tempting move for the second derivative is to divide the second derivatives, just as the first derivatives were divided:

$$\frac{d^2y/dt^2}{d^2x/dt^2} = \frac{-\pi^2\sin(\pi t)}{-\pi^2\cos(\pi t)} = \tan(\pi t)$$

At $t = \frac{1}{4}$ this gives $\tan\frac{\pi}{4} = 1$, a positive number. But $t = \frac{1}{4}$ is the point $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ on the upper half of the circle, and the upper half bends downward like a dome. A concave-down curve must have a negative second derivative, so the shortcut has broken.

The correct method differentiates the first derivative with respect to $t$, then divides by $\frac{dx}{dt}$:

$$\frac{d}{dt}\left(-\cot(\pi t)\right) = \pi\csc^2(\pi t)$$

$$\frac{d^2y}{dx^2} = \frac{\pi\csc^2(\pi t)}{-\pi\sin(\pi t)} = -\csc^3(\pi t)$$

$$\left.\frac{d^2y}{dx^2}\right|_{t = 1/4} = -\left(\sqrt{2}\right)^3 = -2\sqrt{2}$$

Check with implicit differentiation on $x^2 + y^2 = 1$, which gives $\frac{d^2y}{dx^2} = -\frac{1}{y^3}$. At $y = \frac{\sqrt{2}}{2}$ that is $-2\sqrt{2}$ as well.

Final answer: $\frac{dy}{dx} = -\cot(\pi t)$ and $\frac{d^2y}{dx^2} = -\csc^3(\pi t)$, which equals $-2\sqrt{2}$ at $t = \frac{1}{4}$.

Example 4

For the curve $x = 2t^3 - 3t^2$, $y = t^3 - 12t$ (a Brainly question), find every point with a horizontal or vertical tangent.

$$\frac{dx}{dt} = 6t^2 - 6t = 6t(t - 1)$$

$$\frac{dy}{dt} = 3t^2 - 12 = 3(t - 2)(t + 2)$$

Horizontal tangents need $\frac{dy}{dt} = 0$, so $t = 2$ or $t = -2$; at both, $\frac{dx}{dt} = 12$ or $36$, not zero. Vertical tangents need $\frac{dx}{dt} = 0$, so $t = 0$ or $t = 1$; at both, $\frac{dy}{dt} = -12$ or $-9$, not zero.

Final answer: horizontal tangents at $(-28, 16)$ and $(4, -16)$; vertical tangents at $(0, 0)$ and $(-1, -11)$.

Example 5

For the cycloid $x = r(\theta - \sin\theta)$, $y = r(1 - \cos\theta)$, find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$, then evaluate both at $\theta = \frac{\pi}{2}$ when $r = 2$.

$$\frac{dx}{d\theta} = r(1 - \cos\theta), \qquad \frac{dy}{d\theta} = r\sin\theta$$

$$\frac{dy}{dx} = \frac{\sin\theta}{1 - \cos\theta} = \cot\frac{\theta}{2}$$

The last step uses the half-angle identities $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$. For the second derivative:

$$\frac{d}{d\theta}\left(\cot\frac{\theta}{2}\right) = -\frac{1}{2}\csc^2\frac{\theta}{2}$$

$$\frac{d^2y}{dx^2} = \frac{-\frac{1}{2}\csc^2\frac{\theta}{2}}{r(1 - \cos\theta)} = -\frac{1}{r(1 - \cos\theta)^2}$$

At $\theta = \frac{\pi}{2}$ with $r = 2$, the point is $(\pi - 2, 2)$, and:

$$\frac{dy}{dx} = \cot\frac{\pi}{4} = 1, \qquad \frac{d^2y}{dx^2} = -\frac{1}{2(1 - 0)^2} = -\frac{1}{2}$$

The second derivative is negative for every $\theta$ between cusps, so each arch is concave down all the way across. At $\theta = 0, 2\pi, 4\pi, \ldots$ both rates are zero and $\cot\frac{\theta}{2}$ grows without bound, which is the vertical cusp.

Final answer: $\frac{dy}{dx} = \cot\frac{\theta}{2}$ and $\frac{d^2y}{dx^2} = -\frac{1}{r(1 - \cos\theta)^2}$; at the given point the slope is $1$ and the second derivative is $-\frac{1}{2}$.

Example 6

The curve $x = t^2$, $y = t^3 - 3t$ passes through $(3, 0)$ twice. Find both tangent lines there.

Set $y = 0$: $t(t^2 - 3) = 0$, so $t = 0$ or $t = \pm\sqrt{3}$. Only $t = \pm\sqrt{3}$ give $x = 3$, so the curve crosses itself at $(3, 0)$.

$$\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$$

$$t = \sqrt{3}: \quad \frac{9 - 3}{2\sqrt{3}} = \sqrt{3}$$

$$t = -\sqrt{3}: \quad \frac{9 - 3}{-2\sqrt{3}} = -\sqrt{3}$$

Final answer: two tangents at the same point, $y = \sqrt{3}(x - 3)$ and $y = -\sqrt{3}(x - 3)$. A graph $y = f(x)$ can never do this, which is one more reason parametric form exists.

Why Do Curves Need A Parameter In The First Place?

Curves need a parameter because many important paths are motions, and a motion is two coordinates changing with time, not one coordinate depending on the other. The cycloid shows why this mattered historically:

• The curve of quickest descent. In June 1696, Johann Bernoulli challenged mathematicians in Acta Eruditorum to find the path along which a bead slides fastest between two points under gravity. The answer, the brachistochrone curve, is an arc of a cycloid.

• The clock that keeps time on any swing. Christiaan Huygens proved in Horologium Oscillatorium (1673) that a cycloid is tautochronous: a bead released anywhere on it reaches the bottom in the same time.

• Slopes without a formula for $y$. Both results needed the slope and bending of a curve whose $y$ cannot be written as a neat function of $x$. Parametric differentiation is what makes that question answerable.

The same problem returns every time something moves along a path: a ball in flight, a robot arm, a pen on a plotter. The coordinates are easy to write in time; the slope comes from dividing their rates.

Where Are Parametric Derivatives Used In Real Life?

Parametric derivatives turn up wherever a path is designed or tracked, because the slope of the path is the direction of travel.

• Projectiles. A ball launched at speed $v$ and angle $\alpha$ follows $x = v\cos\alpha, t$, $y = v\sin\alpha, t - \frac{g t^2}{2}$. Its slope is $\tan\alpha - \frac{g t}{v\cos\alpha}$, which reaches zero, the top of the flight, at $t = \frac{v\sin\alpha}{g}$.

• Computer fonts and design. The letters on this screen are drawn with Bézier curves, which are parametric. The tangent direction at each end is what lets two pieces join without a visible kink.

• Gears and clockwork. Cycloidal tooth profiles, still used in clocks, are shaped so that the tangent directions of meshing teeth match as they roll.

• Arc length. The length of a parametric path is built from the same two rates, $\frac{dx}{dt}$ and $\frac{dy}{dt}$; see arc length of a curve.

Where Solutions Go Off The Rails

Parametric differentiation has very few moving parts, which makes the errors predictable.

Mistake 1: Dividing The Wrong Way Up

Where it slips in:

The very first line of the problem, usually under time pressure.

Don't do this:

Writing $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$, which gives the reciprocal of the slope.

The correct way:

Keep $y$ on top on both sides: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. The rusher, who writes the two rates down and divides in whatever order they appeared, is the student this catches; saying "the $dt$s cancel" out loud before dividing fixes the order every time.

Mistake 2: Dividing The Second Derivatives

Where it slips in:

Any question that asks for $\frac{d^2y}{dx^2}$ or for concavity.

Don't do this:

Computing $\frac{d^2y/dt^2}{d^2x/dt^2}$. On the circle in Example 3 it gives $+1$ where the true value is $-2\sqrt{2}$.

The correct way:

Differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $\frac{dx}{dt}$. The memorizer who learned "divide the $t$ derivatives" as a rule, rather than as the chain rule, extends it one step too far; the first instinct is to treat the second derivative as the same pattern, and the fix is to remember why the first one worked.

Mistake 3: Stopping One Step Early

Where it slips in:

The second derivative, right after finding $\frac{d}{dt}\left(\frac{dy}{dx}\right)$.

Don't do this:

Reporting $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ as the answer. For the cycloid that would be $-\frac{1}{2}\csc^2\frac{\theta}{2}$, missing the division by $r(1 - \cos\theta)$.

The correct way:

Divide by $\frac{dx}{dt}$ again. A quick check: if $r$ (or any scale) has vanished from a second derivative that should depend on it, a division is missing.

Mistake 4: Calling Every Zero Of $\frac{dx}{dt}$ A Vertical Tangent

Where it slips in:

Tangent-finding questions on curves with cusps or stopping points.

Don't do this:

Listing every $t$ with $\frac{dx}{dt} = 0$ as a vertical tangent without checking $\frac{dy}{dt}$.

The correct way:

Check both rates. If both are zero, find the limit of $\frac{dy}{dx}$ as $t$ approaches that value before naming the tangent. The silent understander usually gets the cycloid cusp right by picture but cannot say why; the limit is the reason.

The habit that prevents most of these is writing the two rates on separate lines before combining them. Students who skip straight to the quotient lose track of which rate is which, and the inverted fraction in Mistake 1 is the usual result.

A wrong curve chosen with confidence has a famous real-world version. In 1638, Galileo argued that a circular arc beats any straight chord for quickest descent, which is true, but the fastest path of all turned out to be the cycloid, found only after Bernoulli's 1696 challenge. Trusting the familiar shape over the calculation is the same error as trusting the familiar-looking division in Mistake 2.

Practice Problems On Derivatives Of A Function In Parametric Form

Work each one, then compare with the answers below. Every answer has been recomputed.

1. Find $\frac{dy}{dx}$ for $x = t^3$, $y = t^2 - 4t$, and evaluate it at $t = 1$.

2. Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ for $x = e^t$, $y = t e^t$.

3. Find the points with horizontal and vertical tangents on $x = 4\cos t$, $y = 3\sin t$.

4. Find $\frac{d^2y}{dx^2}$ for $x = t^2 + 1$, $y = 2t^3$.

5. Find the tangent line to $x = \sec t$, $y = \tan t$ at $t = \frac{\pi}{4}$.

6. Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ for $x = \ln t$, $y = t^2$ with $t > 0$, then confirm by eliminating $t$.

Answer to Question 1: $\frac{dy}{dx} = \frac{2(t - 2)}{3t^2}$, which equals $-\frac{2}{3}$ at $t = 1$.

Answer to Question 2: $\frac{dy}{dx} = 1 + t$ and $\frac{d^2y}{dx^2} = e^{-t}$.

Answer to Question 3: horizontal tangents at $(0, 3)$ and $(0, -3)$; vertical tangents at $(4, 0)$ and $(-4, 0)$.

Answer to Question 4: $\frac{dy}{dx} = 3t$, so $\frac{d^2y}{dx^2} = \frac{3}{2t}$.

Answer to Question 5: the slope is $\csc\frac{\pi}{4} = \sqrt{2}$ at the point $(\sqrt{2}, 1)$, so the tangent is $y = \sqrt{2},x - 1$.

Answer to Question 6: $\frac{dy}{dx} = 2t^2$ and $\frac{d^2y}{dx^2} = 4t^2$; eliminating gives $y = e^{2x}$, with derivatives $2e^{2x}$ and $4e^{2x}$, the same values.

Key Takeaways

• Derivatives of a function in parametric form start from $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, valid wherever $\frac{dx}{dt} \neq 0$.

• The formula is the chain rule solved for the slope, and the slope does not change when the curve is traced faster.

• The second derivative is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ divided by $\frac{dx}{dt}$, never the ratio of the two second derivatives.

• Horizontal tangents come from $\frac{dy}{dt} = 0$, vertical tangents from $\frac{dx}{dt} = 0$, and points where both vanish need a limit.

• Circles, ellipses and cycloids are handled without ever writing $y$ as a function of $x$.

Where To Go From Here

Work through Practice Problems 1 to 6; if Problem 2 or 4 trips you up, return to the second derivative section and Mistake 3, and if Problem 3 does, reread the tangent table above Example 1.

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Read More

• Derivative of a function: the definition every parametric slope rests on.

• Table of derivatives: the trigonometric and exponential derivatives used in each example.

• Differentiation of trigonometric functions: why $\cot$ and $\csc$ keep appearing in circle and cycloid slopes.

• Curve sketching: slope and concavity working together on a graph.

• Related rates: another place where two rates tied to time are combined with the chain rule.