G5PureStretch

Implicit and Parametric Differentiation

Q: Why does differentiating a y-term give an extra dy/dx?

Because y is itself a function of x (even though we don't know its explicit formula). Differentiating y^n needs the chain rule: differentiate the outer power, then multiply by dydx (the derivative of the inner function y with respect to x).

Q: Why is the parametric formula a division and not a multiplication?

By the chain rule, dydx= dydt dtdx. Since dtdx is the reciprocal of dxdt, this becomes dydt dxdt — a division of the two known t-derivatives.

Q: When do I need the product rule in implicit differentiation?

Whenever a term multiplies x and y together, like xy or x^2y^3 — treat it as a product of two functions of x and apply the product rule, remembering the chain rule factor on any y part.

Not every curve can be rearranged into " $y =$ something" — the equation of a circle mixes $x$ and $y$ together inextricably, and a parametric curve gives both $x$ and $y$ in terms of a third variable. Both situations need a differentiation technique beyond the standard rules.

34 min Video by TLMaths Differentiation

The big picture

So far every function has been $explicit$ : $y$ written directly in terms of $x$ . Many important curves — circles, ellipses, and countless real relationships — are only given $implicitly$ , with $x$ and $y$ tangled together, or $parametrically$ , with both defined via a parameter $t$ . This lesson extends the chain rule to both situations, so that literally any curve, however it is presented, can be differentiated.

A-Level Maths G5-01: Introducing Implicit and Explicit EquationsWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Differentiate an implicit relation using the chain rule on y-terms
Find $\frac{d y}{d x}$ for an implicit equation and evaluate it at a point
Differentiate parametric equations using $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$
Find the equation of a tangent to a parametric curve

Implicit differentiation

An $implicit$ equation mixes $x$ and $y$ together, like $x^{2} + y^{2} = 25$ (a circle). To differentiate, treat $y$ as a function of $x$ and differentiate EVERY term with respect to $x$ — any term containing $y$ picks up a $\frac{d y}{d x}$ factor from the chain rule, since $y$ is itself a function of $x$ .

\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}

The chain rule applied to any y-term.

1Differentiate every term with respect to

x

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (25)

2 x + 2 y \frac{d y}{d x} = 0

3Rearrange for

\frac{d y}{d x}

2 y \frac{d y}{d x} = - 2 x

, so

\frac{d y}{d x} = - \frac{x}{y}

Answer

\frac{d y}{d x} = - \frac{x}{y}

Tip — A term with BOTH $x$ and $y$ multiplied together (like $3 x y$ or $x^{2} y$ ) always needs the PRODUCT rule as well as the chain rule — this is the step students most often miss.

Parametric differentiation

A $parametric$ curve gives both $x$ and $y$ as separate functions of a third variable, the parameter $t$ : $x = f (t)$ , $y = g (t)$ . To find $\frac{d y}{d x}$ , differentiate each with respect to $t$ separately, then divide.

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Divide the two separate t-derivatives.

\frac{d x}{d t} = 2 t

\frac{d y}{d t} = 6 t^{2}

2Divide:

\frac{d y}{d x} = \frac{6 t ^{2}}{2 t} = 3 t

Answer

\frac{d y}{d x} = 3 t

Why this works

This formula is really just the chain rule in disguise: $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{d y}{d t} \times \frac{1}{d x / d t}$ , which rearranges to the division shown above.

Tangents to a parametric curve

Once $\frac{d y}{d x}$ is known in terms of $t$ , substitute the specific $t$ -value of interest to get a numerical gradient, then find the Cartesian point $(x, y)$ by substituting that same $t$ into $x = f (t)$ and $y = g (t)$ — the tangent equation follows exactly as before.

1Gradient at

t = 1

\frac{d y}{d x} = 3 t = 3

2Point:

x = 1^{2} = 1

y = 2 (1)^{3} = 2

, so

(1, 2)

3Tangent:

y - 2 = 3 (x - 1)

, so

y = 3 x - 1

Answer

y = 3 x - 1

Think like an examiner

For implicit differentiation, every $y$ -term picks up a $\frac{d y}{d x}$ factor — collect ALL such terms on one side before factorising to isolate $\frac{d y}{d x}$ .
Watch for mixed terms like $x y$ or $x^{2} y$ — these need the product rule in addition to the chain rule.
For parametric curves, always differentiate $x$ and $y$ SEPARATELY with respect to $t$ first — never try to eliminate $t$ before differentiating unless specifically required.

Common misconceptions

Differentiating

y^{2}

as simply

2 y

, forgetting the chain rule factor.

Since

y

is a function of

x

\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}

— the

\frac{d y}{d x}

must always be included.

Forgetting the product rule on a mixed term like

x y

x y

is a product of two functions of

x

(treating

y

as one of them), so

\frac{d}{d x} (x y) = x \frac{d y}{d x} + y

Writing

\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d x}{d t}

for parametric curves.

It is DIVISION, not multiplication:

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Implicit and parametric differentiation

\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}

Chain rule on any y-term.

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Parametric differentiation.

Stretch yourself

Find $\frac{d y}{d x}$ at the point $(3, 4)$ on the circle $x^{2} + y^{2} = 25$ , and hence find the gradient of the tangent there.

Hint — Differentiate implicitly to get a general formula for $\frac{d y}{d x}$ , then substitute the given point directly.

Questions students ask

Why does differentiating a y-term give an extra dy/dx?

Because

y

is itself a function of

x

(even though we don't know its explicit formula). Differentiating

y^{n}

needs the chain rule: differentiate the outer power, then multiply by

\frac{d y}{d x}

(the derivative of the inner function

y

with respect to

x

Why is the parametric formula a division and not a multiplication?

By the chain rule,

\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}

. Since

\frac{d t}{d x}

is the reciprocal of

\frac{d x}{d t}

, this becomes

\frac{d y}{d t} \div \frac{d x}{d t}

— a division of the two known

t

-derivatives.

When do I need the product rule in implicit differentiation?

Whenever a term multiplies

x

and

y

together, like

x y

x^{2} y^{3}

— treat it as a product of two functions of

x

and apply the product rule, remembering the chain rule factor on any

y

part.

Key takeaways

Implicit differentiation: differentiate every term with respect to $x$ ; a y-term of power $n$ gives $n y^{n - 1} \frac{d y}{d x}$ .
Mixed terms like $x y$ need the product rule as well as the chain rule.
Collect all $\frac{d y}{d x}$ terms and factorise to isolate it.
Parametric differentiation: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ , found by differentiating $x$ and $y$ separately with respect to $t$ .

How this fits the course

Build on

Integrating xⁿ

Leads to

Integrating xⁿ

Test yourself

Ready to lock in Implicit and Parametric Differentiation? Pick a mode and earn XP & Dobloons.

Implicit and Parametric Differentiation

Not every curve can be rearranged into "

y =

something" — the equation of a circle mixes

x

and

y

together inextricably, and a parametric curve gives both

x

and

y

in terms of a third variable. Both situations need a differentiation technique beyond the standard rules.

34 min Video by TLMaths Differentiation