Mathematics 9709 · Pure Mathematics 2

Parametric Differentiation

Sometimes $x$ and $y$ are both written in terms of a third letter $t$ (or $\theta$ ), e.g. $x=t^2,\ y=5-2t$ .
To get the gradient you go through $t$ : the chain rule one more time.

$t$ is the hidden inside. $\dfrac{dy}{dx}=\dfrac{\,dy/dt\,}{\,dx/dt\,}$ . So differentiate $y$ with respect to $t$ , differentiate $x$ with respect to $t$ , then divide. The answer comes out in $t$ , not $x$ .

The one-line method

(Worked Example 4.14)

Worked example

$x = t^2,\ y = 5 - 2t$ : find $\dfrac{dy}{dx}$ , then the tangent at $t = 2$

Differentiate each with respect to $t$ : $\dfrac{dx}{dt}=2t$ , $\dfrac{dy}{dt}=-2$ .
Divide ( $y$ on top, like $\frac{dy}{dx}$ ): $\dfrac{dy}{dx}=\dfrac{-2}{2t}=-\dfrac1t$ .
At $t=2$ : gradient $-\tfrac12$ , point $(t^2,\,5-2t)=(4,\,1)$ .
Line $y-y_1=m(x-x_1)$ : $y-1=-\tfrac12(x-4)$ .

Answer

y=-\tfrac12 x+3

Worked example

Real question: $x = t e^{2t},\ y = t^2 + t + 3$ , show $\dfrac{dy}{dx} = e^{-2t}$

9709/32 F/M 2023 Q5(a). The product rule on $x$ leads to a clean cancellation, and the whole question rests on it.

$x=te^{2t}$ is a product (inside derivative $2$ ): $\dfrac{dx}{dt}=t(2e^{2t})+e^{2t}=(2t+1)e^{2t}$ .
$y=t^2+t+3$ : $\dfrac{dy}{dt}=2t+1$ .
Divide, and the $(2t+1)$ cancels top and bottom: $\dfrac{dy}{dx}=\dfrac{2t+1}{(2t+1)e^{2t}}=\dfrac{1}{e^{2t}}$ .

Answer

\dfrac{dy}{dx}=e^{-2t}

Spotting the shared $(2t+1)$ factor turns a messy answer into one line. That cancellation is the whole question. (9709/32 Feb/March 2023 Q5)

Common mistake

Three parametric traps to avoid:

Inverting the fraction — writing $\frac{dx/dt}{dy/dt}$ . Memory hook: $y$ on top, just like $\frac{dy}{dx}$ .
Leaving the answer in $t$ when a numeric gradient or point is asked — substitute the given $t$ .
Axis confusion (see below) — mixing up which derivative is zero.

'Show that' and tangent parts

On a “show that $\frac{dy}{dx}=\dots$ ”, the tidy answer is already given. So you must show every algebra step and every trig-identity step. The target form tells you which identity to use.

Worked example

Real question: $x = e^{\tan t},\ y = 3\tan^2 t$ , tangent at $(e, 3)$

9709/31 M/J 2025 Q4. Find the gradient through $t$ , then write an ordinary P1 tangent line with exact $m$ and $c$ .

Differentiate (both need the chain rule, inside $\tan t$ with derivative $\sec^2 t$ ): $\dfrac{dx}{dt}=\sec^2 t\,e^{\tan t}$ , $\dfrac{dy}{dt}=6\tan t\sec^2 t$ .
Divide; the $\sec^2 t$ cancels: $\dfrac{dy}{dx}=\dfrac{6\tan t}{e^{\tan t}}=6\tan t\,e^{-\tan t}$ .
At $(e,3)$ , $x=e^{\tan t}=e$ gives $\tan t=1$ , so $m=6(1)e^{-1}=\dfrac{6}{e}$ .
Line through $(e,3)$ : $y-3=\dfrac{6}{e}(x-e)=\dfrac{6}{e}x-6$ .

Answer

y=\dfrac{6}{e}x-3

Reading $\tan t=1$ off the given point is the step that gives you an exact gradient. Never reach for a decimal. (9709/31 May/June 2025 Q4)

If the divided fraction comes out as $\dfrac{\cos t}{\sin t}$ , a target of $\cot t$ is your hint to write it as a single cotangent.
These parts often mix in logs and exponentials. For example, $x=t+3,\ y=\ln t$ gives $\frac{dy}{dx}=\frac1t$ . (9709/31 Oct/Nov 2023 Q6)

Tangents parallel to an axis

Parallel to the $x$ -axis ⇒ gradient $0$ ⇒ set the top $\frac{dy}{dt}=0$ . Parallel to the $y$ -axis ⇒ straight up, infinite gradient ⇒ set the bottom $\frac{dx}{dt}=0$ .

The why: a flat tangent needs the up-down rate to stop, a vertical tangent needs the side-to-side rate to stop. (Worked Example 4.15)

Your turn— tap to reveal the worked answer (9709-style: x = 3t, y = t³ + 4t² − 3t, gradient 0)

$\frac{dx}{dt}=3$ , $\frac{dy}{dt}=3t^2+8t-3$ , so $\frac{dy}{dx}=\frac{3t^2+8t-3}{3}$ . Gradient $0$ when the top is zero: $3t^2+8t-3=0\Rightarrow(3t-1)(t+3)=0$ .

Answer

t=\tfrac13\ \text{ or }\ t=-3

Maximum-gradient and second-derivative follow-ups sit in the advanced layer.

PreviousImplicit Differentiation NextIntegration of exponential functions

Parametric Differentiation

Sometimes $x$ and $y$ are both written in terms of a third letter $t$ (or $\theta$ ), e.g. $x=t^2,\ y=5-2t$ .
To get the gradient you go through $t$ : the chain rule one more time.

The one-line method

(Worked Example 4.14)

Worked example

$x = t^2,\ y = 5 - 2t$ : find $\dfrac{dy}{dx}$ , then the tangent at $t = 2$

Differentiate each with respect to $t$ : $\dfrac{dx}{dt}=2t$ , $\dfrac{dy}{dt}=-2$ .
Divide ( $y$ on top, like $\frac{dy}{dx}$ ): $\dfrac{dy}{dx}=\dfrac{-2}{2t}=-\dfrac1t$ .
At $t=2$ : gradient $-\tfrac12$ , point $(t^2,\,5-2t)=(4,\,1)$ .
Line $y-y_1=m(x-x_1)$ : $y-1=-\tfrac12(x-4)$ .

Answer

y=-\tfrac12 x+3

Worked example

Real question: $x = t e^{2t},\ y = t^2 + t + 3$ , show $\dfrac{dy}{dx} = e^{-2t}$

9709/32 F/M 2023 Q5(a). The product rule on $x$ leads to a clean cancellation, and the whole question rests on it.

$x=te^{2t}$ is a product (inside derivative $2$ ): $\dfrac{dx}{dt}=t(2e^{2t})+e^{2t}=(2t+1)e^{2t}$ .
$y=t^2+t+3$ : $\dfrac{dy}{dt}=2t+1$ .
Divide, and the $(2t+1)$ cancels top and bottom: $\dfrac{dy}{dx}=\dfrac{2t+1}{(2t+1)e^{2t}}=\dfrac{1}{e^{2t}}$ .

Answer

\dfrac{dy}{dx}=e^{-2t}

Spotting the shared $(2t+1)$ factor turns a messy answer into one line. That cancellation is the whole question. (9709/32 Feb/March 2023 Q5)

Common mistake

Three parametric traps to avoid:

Inverting the fraction — writing $\frac{dx/dt}{dy/dt}$ . Memory hook: $y$ on top, just like $\frac{dy}{dx}$ .
Leaving the answer in $t$ when a numeric gradient or point is asked — substitute the given $t$ .
Axis confusion (see below) — mixing up which derivative is zero.

'Show that' and tangent parts

Worked example

Real question: $x = e^{\tan t},\ y = 3\tan^2 t$ , tangent at $(e, 3)$

9709/31 M/J 2025 Q4. Find the gradient through $t$ , then write an ordinary P1 tangent line with exact $m$ and $c$ .

Differentiate (both need the chain rule, inside $\tan t$ with derivative $\sec^2 t$ ): $\dfrac{dx}{dt}=\sec^2 t\,e^{\tan t}$ , $\dfrac{dy}{dt}=6\tan t\sec^2 t$ .
Divide; the $\sec^2 t$ cancels: $\dfrac{dy}{dx}=\dfrac{6\tan t}{e^{\tan t}}=6\tan t\,e^{-\tan t}$ .
At $(e,3)$ , $x=e^{\tan t}=e$ gives $\tan t=1$ , so $m=6(1)e^{-1}=\dfrac{6}{e}$ .
Line through $(e,3)$ : $y-3=\dfrac{6}{e}(x-e)=\dfrac{6}{e}x-6$ .

Answer

y=\dfrac{6}{e}x-3

Reading $\tan t=1$ off the given point is the step that gives you an exact gradient. Never reach for a decimal. (9709/31 May/June 2025 Q4)

If the divided fraction comes out as $\dfrac{\cos t}{\sin t}$ , a target of $\cot t$ is your hint to write it as a single cotangent.
These parts often mix in logs and exponentials. For example, $x=t+3,\ y=\ln t$ gives $\frac{dy}{dx}=\frac1t$ . (9709/31 Oct/Nov 2023 Q6)

Tangents parallel to an axis

Parallel to the $x$ -axis ⇒ gradient $0$ ⇒ set the top $\frac{dy}{dt}=0$ . Parallel to the $y$ -axis ⇒ straight up, infinite gradient ⇒ set the bottom $\frac{dx}{dt}=0$ .

The why: a flat tangent needs the up-down rate to stop, a vertical tangent needs the side-to-side rate to stop. (Worked Example 4.15)

Your turn— tap to reveal the worked answer (9709-style: x = 3t, y = t³ + 4t² − 3t, gradient 0)

$\frac{dx}{dt}=3$ , $\frac{dy}{dt}=3t^2+8t-3$ , so $\frac{dy}{dx}=\frac{3t^2+8t-3}{3}$ . Gradient $0$ when the top is zero: $3t^2+8t-3=0\Rightarrow(3t-1)(t+3)=0$ .

Answer

t=\tfrac13\ \text{ or }\ t=-3

Maximum-gradient and second-derivative follow-ups sit in the advanced layer.