Parametric derivatives worksheet

With x and y both given in terms of t, dy/dx is dy/dt divided by dx/dt. The second derivative is the derivative of that ratio with respect to t, divided by dx/dt once more.

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Sigma Prep · sigmaprep.io/worksheets Name Parametric Derivatives Date Period Answer each question about the curve. $x = t^{2}$ , $y = 2 t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $x = t^{2} + 2 t$ , $y = t^{3} - 27 t$ . For what $t$ is the tangent line vertical? $x = 2 t^{2} - 2 t$ , $y = t^{3} - 3 t - 3$ . Find the points where the tangent line is horizontal. $x = - 3 t^{2} - 2 t$ , $y = - t^{3} - t$ . Find $\frac{d y}{d x}$ at $t = 3$ $x = t^{2} + 2 t$ , $y = t^{3} - 3 t^{2} + 2$ . For what $t$ is the tangent line horizontal? $x = t^{2} - 4 t$ , $y = t^{3} - 27 t + 2$ . Find the points where the tangent line is vertical. $x = - 3 t^{2} - 3 t$ , $y = - t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = - 3$ $x = t^{2} - 2 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line vertical? $x = 3 t^{2} + 3 t$ , $y = t^{3} - 12 t - 3$ . Find the points where the tangent line is horizontal. $x = - 3 t^{2} + 4 t$ , $y = 2 t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $x = t^{2} + 6 t$ , $y = t^{3} - 9 t^{2} + 24 t - 18$ . For what $t$ is the tangent line vertical? $x = t^{2} - 2 t$ , $y = t^{3} - 27 t + 5$ . Find the points where the tangent line is horizontal. $x = - t^{2} + t$ , $y = 2 t^{3} - 6 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $x = t^{2} - 6 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line horizontal? $x = t^{2} - t$ , $y = t^{3} - 12 t - 4$ . Find the points where the tangent line is horizontal. $x = 2 t^{2} - 3 t$ , $y = 2 t^{3} - 4 t$ . Find $\frac{d y}{d x}$ at $t = 1$ $x = t^{2} - 6 t$ , $y = t^{3} - 9 t^{2} + 54$ . For what $t$ is the tangent line vertical? $x = t^{2} + 4 t$ , $y = t^{3} - 3 t + 4$ . Find the points where the tangent line is horizontal. $x = - t^{2} - t$ , $y = t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = - 3$ $x = t^{2} + 4 t$ , $y = t^{3} + 3 t^{2} - 9 t - 11$ . For what $t$ is the tangent line vertical? Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

Parametric Derivatives

Date Period

Answer each question about the curve.

$x = t^{2}$ , $y = 2 t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = 3$
$x = t^{2} + 2 t$ , $y = t^{3} - 27 t$ . For what $t$ is the tangent line vertical?
$x = 2 t^{2} - 2 t$ , $y = t^{3} - 3 t - 3$ . Find the points where the tangent line is horizontal.
$x = - 3 t^{2} - 2 t$ , $y = - t^{3} - t$ . Find $\frac{d y}{d x}$ at $t = 3$
$x = t^{2} + 2 t$ , $y = t^{3} - 3 t^{2} + 2$ . For what $t$ is the tangent line horizontal?
$x = t^{2} - 4 t$ , $y = t^{3} - 27 t + 2$ . Find the points where the tangent line is vertical.
$x = - 3 t^{2} - 3 t$ , $y = - t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = - 3$
$x = t^{2} - 2 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line vertical?
$x = 3 t^{2} + 3 t$ , $y = t^{3} - 12 t - 3$ . Find the points where the tangent line is horizontal.
$x = - 3 t^{2} + 4 t$ , $y = 2 t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = 3$
$x = t^{2} + 6 t$ , $y = t^{3} - 9 t^{2} + 24 t - 18$ . For what $t$ is the tangent line vertical?
$x = t^{2} - 2 t$ , $y = t^{3} - 27 t + 5$ . Find the points where the tangent line is horizontal.
$x = - t^{2} + t$ , $y = 2 t^{3} - 6 t$ . Find $\frac{d y}{d x}$ at $t = 3$
$x = t^{2} - 6 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line horizontal?
$x = t^{2} - t$ , $y = t^{3} - 12 t - 4$ . Find the points where the tangent line is horizontal.
$x = 2 t^{2} - 3 t$ , $y = 2 t^{3} - 4 t$ . Find $\frac{d y}{d x}$ at $t = 1$
$x = t^{2} - 6 t$ , $y = t^{3} - 9 t^{2} + 54$ . For what $t$ is the tangent line vertical?
$x = t^{2} + 4 t$ , $y = t^{3} - 3 t + 4$ . Find the points where the tangent line is horizontal.
$x = - t^{2} - t$ , $y = t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = - 3$
$x = t^{2} + 4 t$ , $y = t^{3} + 3 t^{2} - 9 t - 11$ . For what $t$ is the tangent line vertical?



Sigma Prep · sigmaprep.io/worksheets Name Parametric DerivativesAnswer keyVersion 1 Date Period Answer each question about the curve. $x = t^{2}$ , $y = 2 t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $\frac{17}{2}$ $x = t^{2} + 2 t$ , $y = t^{3} - 27 t$ . For what $t$ is the tangent line vertical? $t = - 1$ $x = 2 t^{2} - 2 t$ , $y = t^{3} - 3 t - 3$ . Find the points where the tangent line is horizontal. $(4, - 1)$ and $(0, - 5)$ $x = - 3 t^{2} - 2 t$ , $y = - t^{3} - t$ . Find $\frac{d y}{d x}$ at $t = 3$ $\frac{7}{5}$ $x = t^{2} + 2 t$ , $y = t^{3} - 3 t^{2} + 2$ . For what $t$ is the tangent line horizontal? $t = 0$ , $t = 2$ $x = t^{2} - 4 t$ , $y = t^{3} - 27 t + 2$ . Find the points where the tangent line is vertical. $(- 4, - 44)$ $x = - 3 t^{2} - 3 t$ , $y = - t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = - 3$ $- 2$ $x = t^{2} - 2 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line vertical? $t = 1$ $x = 3 t^{2} + 3 t$ , $y = t^{3} - 12 t - 3$ . Find the points where the tangent line is horizontal. $(6, 13)$ and $(18, - 19)$ $x = - 3 t^{2} + 4 t$ , $y = 2 t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $- \frac{59}{14}$ $x = t^{2} + 6 t$ , $y = t^{3} - 9 t^{2} + 24 t - 18$ . For what $t$ is the tangent line vertical? $t = - 3$ $x = t^{2} - 2 t$ , $y = t^{3} - 27 t + 5$ . Find the points where the tangent line is horizontal. $(15, 59)$ and $(3, - 49)$ $x = - t^{2} + t$ , $y = 2 t^{3} - 6 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $- \frac{48}{5}$ $x = t^{2} - 6 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line horizontal? $t = - 3$ , $t = - 1$ $x = t^{2} - t$ , $y = t^{3} - 12 t - 4$ . Find the points where the tangent line is horizontal. $(6, 12)$ and $(2, - 20)$ $x = 2 t^{2} - 3 t$ , $y = 2 t^{3} - 4 t$ . Find $\frac{d y}{d x}$ at $t = 1$ $2$ $x = t^{2} - 6 t$ , $y = t^{3} - 9 t^{2} + 54$ . For what $t$ is the tangent line vertical? $t = 3$ $x = t^{2} + 4 t$ , $y = t^{3} - 3 t + 4$ . Find the points where the tangent line is horizontal. $(- 3, 6)$ and $(5, 2)$ $x = - t^{2} - t$ , $y = t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = - 3$ $\frac{32}{5}$ $x = t^{2} + 4 t$ , $y = t^{3} + 3 t^{2} - 9 t - 11$ . For what $t$ is the tangent line vertical? $t = - 2$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

Parametric DerivativesAnswer keyVersion 1

Date Period

Answer each question about the curve.

$x = t^{2}$ , $y = 2 t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $\frac{17}{2}$
$x = t^{2} + 2 t$ , $y = t^{3} - 27 t$ . For what $t$ is the tangent line vertical? $t = - 1$
$x = 2 t^{2} - 2 t$ , $y = t^{3} - 3 t - 3$ . Find the points where the tangent line is horizontal. $(4, - 1)$ and $(0, - 5)$
$x = - 3 t^{2} - 2 t$ , $y = - t^{3} - t$ . Find $\frac{d y}{d x}$ at $t = 3$ $\frac{7}{5}$
$x = t^{2} + 2 t$ , $y = t^{3} - 3 t^{2} + 2$ . For what $t$ is the tangent line horizontal? $t = 0$ , $t = 2$
$x = t^{2} - 4 t$ , $y = t^{3} - 27 t + 2$ . Find the points where the tangent line is vertical. $(- 4, - 44)$
$x = - 3 t^{2} - 3 t$ , $y = - t^{3} - 3 t$ . Find $\frac{d y}{d x}$ at $t = - 3$ $- 2$
$x = t^{2} - 2 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line vertical? $t = 1$
$x = 3 t^{2} + 3 t$ , $y = t^{3} - 12 t - 3$ . Find the points where the tangent line is horizontal. $(6, 13)$ and $(18, - 19)$
$x = - 3 t^{2} + 4 t$ , $y = 2 t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $- \frac{59}{14}$
$x = t^{2} + 6 t$ , $y = t^{3} - 9 t^{2} + 24 t - 18$ . For what $t$ is the tangent line vertical? $t = - 3$
$x = t^{2} - 2 t$ , $y = t^{3} - 27 t + 5$ . Find the points where the tangent line is horizontal. $(15, 59)$ and $(3, - 49)$
$x = - t^{2} + t$ , $y = 2 t^{3} - 6 t$ . Find $\frac{d y}{d x}$ at $t = 3$ $- \frac{48}{5}$
$x = t^{2} - 6 t$ , $y = t^{3} + 6 t^{2} + 9 t + 2$ . For what $t$ is the tangent line horizontal? $t = - 3$ , $t = - 1$
$x = t^{2} - t$ , $y = t^{3} - 12 t - 4$ . Find the points where the tangent line is horizontal. $(6, 12)$ and $(2, - 20)$
$x = 2 t^{2} - 3 t$ , $y = 2 t^{3} - 4 t$ . Find $\frac{d y}{d x}$ at $t = 1$ $2$
$x = t^{2} - 6 t$ , $y = t^{3} - 9 t^{2} + 54$ . For what $t$ is the tangent line vertical? $t = 3$
$x = t^{2} + 4 t$ , $y = t^{3} - 3 t + 4$ . Find the points where the tangent line is horizontal. $(- 3, 6)$ and $(5, 2)$
$x = - t^{2} - t$ , $y = t^{3} + 5 t$ . Find $\frac{d y}{d x}$ at $t = - 3$ $\frac{32}{5}$
$x = t^{2} + 4 t$ , $y = t^{3} + 3 t^{2} - 9 t - 11$ . For what $t$ is the tangent line vertical? $t = - 2$

How to do these

$x = t^{2}$ , $y = t^{3}$ . Find $\frac{d y}{d x}$ at $t = 2$

dy/dt = 3t² = 12 and dx/dt = 2t = 4.
dy/dx = 12/4.

The answer is $3$ .

Where students go wrong

Finding d²y/dx² by differentiating dy/dx with respect to t and stopping there. That is a rate per unit of t; it still has to be divided by dx/dt.

Also called parametric differentiation, dy/dx parametric or second derivative parametric.

What you can put on this worksheet

dy/dx and the second derivative: $x = 2 t^{2} + 2 t$ , $y = - t^{3} + t$ . Find $\frac{d y}{d x}$ at $t = - 3$ → $\frac{13}{5}$
Horizontal and vertical tangents: the value of t: $x = t^{2}$ , $y = t^{3} - 3 t^{2} - 9 t + 11$ . For what $t$ is the tangent line horizontal? → $t = - 1$ , $t = 3$
Horizontal and vertical tangents: the points: $x = 2 t^{2}$ , $y = t^{3} - 12 t + 3$ . Find the points where the tangent line is horizontal. → $(8, 19)$ and $(8, - 13)$

Questions about these worksheets

Yes. Every parametric derivatives sheet is free to print and free to download as a PDF. There is no account to make, no email to hand over and no limit on how many you take.

Yes. The answer key prints on a second page, and you can choose whether it shows each question beside its answer or just the answers in a list.

Yes. Print straight from the page, or download the PDF and print that. The sheet is laid out for paper rather than squeezed off a screen, so there is room to work under each question, and the answer key comes out on its own page.

Calculus is usually taken around 12th grade or a first college course, though schools vary and plenty of students meet it earlier or later. Pick the difficulty rather than the grade: the easy level suits a first lesson on it, the hard level suits review before a test.

Yes, as many as you like. The questions are built fresh each time rather than picked from a fixed set of files, so pressing Generate gives a new sheet. That is what you want for a second class, a retake, or two students sitting next to each other.

That is the closest comparison, and Kuta Software is good software. It is also paid software you install on a computer. These worksheets run in a browser tab for free, with no account and nothing to install. Like Kuta, the parametric derivatives questions are generated when you ask for them rather than pulled from a fixed set of files, so two classes never get the same sheet, and the answer key comes with it.

Yes. There are three levels, and you can tick more than one for a sheet that mixes them, in which case the questions come out easiest first. The harder levels are not just larger numbers: they ask for something the easy ones do not.

You pick from 3: dy/dx and the second derivative, horizontal and vertical tangents: the value of t and horizontal and vertical tangents: the points. Tick as many as you want and set how many of each, or let it spread them evenly.

Finding d²y/dx² by differentiating dy/dx with respect to t and stopping there. That is a rate per unit of t; it still has to be divided by dx/dt.

Parametric derivatives worksheet

Select your difficulty

Parametric Derivatives

Parametric DerivativesAnswer keyVersion 1

How to do these

Where students go wrong

What you can put on this worksheet

Questions about these worksheets

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