9709 P3 §3.4 Differentiation — Lesson 6

Implicit Differentiation
A-Level Mathematics · §3.4 Differentiation
💡 Lead In

The Circle's Real Slope

On the circle $x^{2}+y^{2}=25$, a student differentiates both sides and claims the gradient at $(3,4)$ is $-3$. Diagnose each line: click Correct step or Wrong step.

Step 1 — $\dfrac{d}{dx}\big(x^{2}+y^{2}\big)=2x+2y$
Step 2 — $2x+2y=0$, so $\dfrac{dy}{dx}=-x=-3$ at $(3,4)$
Step 3 — corrected: $2x+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac{3}{4}$

The slip: $y^{2}$ differentiates to $2y\dfrac{dy}{dx}$, not $2y$ — the chain rule applies to every $y$-term. True gradient: $-\dfrac{3}{4}$.

🧠 Activating Prior Knowledge

Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.

R1. If $y$ is a function of $x$, $\dfrac{d}{dx}\big(y^{3}\big)$ = ?

$3y^{2}\dfrac{dy}{dx}$
$3y^{2}$
$y^{3}\dfrac{dy}{dx}$
$3y^{2}x$
Choose an option above.
Chain rule: the outer power $3y^{2}$ times the inner derivative $\dfrac{dy}{dx}$.

R2. $\dfrac{d}{dx}\big(x^{2}\big)$ = ?

$2$
$2x$
$x^{2}$
$x$
Choose an option above.
Power rule: $2x$.

R3. The product rule on $xy$ gives $\dfrac{d}{dx}(xy)$ = ?

$y$
$1$
$y+x\dfrac{dy}{dx}$
$x\dfrac{dy}{dx}$
Choose an option above.
$u=x,\ v=y$: $u'v+uv'=y+x\dfrac{dy}{dx}$.
📖 Key Concepts
Explore 1 — differentiate both sides, keeping every $y$ alive.

For $x^{2}+y^{2}=25$, treat $y$ as a function of $x$. Every $y$-power is a composite whose inner derivative is $\dfrac{dy}{dx}$:

$2x+2y\dfrac{dy}{dx}=0\ \Rightarrow\ \dfrac{dy}{dx}=-\dfrac{x}{y}$

The chain rule from Lesson 2 is the engine: the derivative of $y^{n}$ always carries a factor $\dfrac{dy}{dx}$.

Explore 2 — product terms and solving for $\dfrac{dy}{dx}$.
$xy+x+y=5$

$y+x\dfrac{dy}{dx}+1+\dfrac{dy}{dx}=0$ — the term $xy$ needs the product rule from Lesson 3.

Collect the $\dfrac{dy}{dx}$ terms: $(x+1)\dfrac{dy}{dx}=-(y+1)$, so $\dfrac{dy}{dx}=-\dfrac{y+1}{x+1}$.

Always gather every $\dfrac{dy}{dx}$ on one side before solving.

Explore 3 — the chain and product rules working together.
$x^{2}+3xy+y^{2}=11$

$2x+3y+3x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{2x+3y}{3x+2y}$

The $3xy$ term contributes both $3y$ (from $u'v$) and $3x\dfrac{dy}{dx}$ (from $uv'$).

Group Match the Lines: implicit curve → dy/dx

Groups of 4: tap an implicit equation on the left, then its dy/dx on the right — a coloured line links each pair. Six curves covering circles, products and mixed powers. Wrong pairs turn red to fix.

Individual Spot the Error: implicit slips

Each attempt below claims a dy/dx. Decide whether implicit differentiation was applied correctly, then explain your verdict out loud to a partner.

✏️ Worked Examples
Exercise 4D Q1(ii) Worked Example

Given $x^{2}+y^{3}=8$, find $\dfrac{dy}{dx}$.

Step 1 — differentiate both sides

$2x+3y^{2}\dfrac{dy}{dx}=0$.

Step 2 — solve for $\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=-\dfrac{2x}{3y^{2}}$.

Exercise 4D Q4 Worked Example

Find the gradient of $x^{2}+3xy+y^{2}=x+3y$ at $(2,-1)$, and the equation of the tangent there.

Step 1 — differentiate both sides

$2x+3y+3x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=1+3\dfrac{dy}{dx}$.

Step 2 — solve for $\dfrac{dy}{dx}$

$(3x+2y-3)\dfrac{dy}{dx}=1-2x-3y$, so $\dfrac{dy}{dx}=\dfrac{1-2x-3y}{3x+2y-3}$.

Step 3 — evaluate at $(2,-1)$

$\dfrac{dy}{dx}=\dfrac{1-4+3}{6-2-3}=\dfrac{0}{1}=0$. The tangent is horizontal: $y=-1$.

📝 Concept Quiz

Answer, then check. Options are shuffled every load. The clock starts on your first answer.

Q1. $\dfrac{d}{dx}\big(y^{3}\big)$ = ?

$3y^{2}$
$3y^{2}\dfrac{dy}{dx}$
$3y^{3}\dfrac{dy}{dx}$
$\tfrac14 y^{4}$
Choose an option above.
Chain rule on the $y$-power: $3y^{2}\dfrac{dy}{dx}$.

Q2. $\dfrac{d}{dx}(xy)$ = ?

$y+x\dfrac{dy}{dx}$
$y$
$x\dfrac{dy}{dx}$
$1$
Choose an option above.
Product rule: $u'v+uv'=y+x\dfrac{dy}{dx}$.

Q3. For $x^{2}+y^{2}=10$, $\dfrac{dy}{dx}$ = ?

$\dfrac{x}{y}$
$-\dfrac{y}{x}$
$-\dfrac{x}{y}$
$-\dfrac{x}{2y}$
Choose an option above.
$2x+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{x}{y}$.

Q4. The gradient of $x^{2}+y^{2}=25$ at $(3,4)$ is:

$-\dfrac{3}{4}$
$-3$
$-\dfrac{4}{3}$
$\dfrac{3}{4}$
Choose an option above.
$\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac{3}{4}$ at $(3,4)$.
📚 More Practice

Work each on paper, then reveal the solution to check your working.

Exercise 4D Q1(i) Standard

Differentiate $y^{4}$ with respect to $x$.

Reveal Solution

$\dfrac{d}{dx}\big(y^{4}\big)=4y^{3}\dfrac{dy}{dx}$.

Exercise 4D Q1(x) Standard

Differentiate $x^{2}\ln y$ with respect to $x$.

Reveal Solution

Product rule: $2x\ln y+x^{2}\cdot\dfrac{1}{y}\dfrac{dy}{dx}=2x\ln y+\dfrac{x^{2}}{y}\dfrac{dy}{dx}$.

Exercise 4D Q2 Medium

Find the gradient of $xy^{3}=5\ln y$ at $(0,1)$.

Reveal Solution

$y^{3}+3xy^{2}\dfrac{dy}{dx}=\dfrac{5}{y}\dfrac{dy}{dx}$. At $(0,1)$: $1=5\dfrac{dy}{dx}$, so $\dfrac{dy}{dx}=\dfrac{1}{5}$.

Exercise 4D Q5 Hard

Find the coordinates of all the stationary points on the curve $x^{2}+y^{2}+xy=3$.

Reveal Solution

$\dfrac{dy}{dx}=-\dfrac{2x+y}{2y+x}$. Stationary when $2x+y=0$, i.e. $y=-2x$.

Substitute: $x^{2}+4x^{2}-2x^{2}=3$, so $3x^{2}=3$ and $x=\pm1$.

Points $(1,-2)$ and $(-1,2)$. The second derivative shows $(1,-2)$ is a minimum and $(-1,2)$ is a maximum.

🎫

Exit Ticket

Check the learning objectives you feel confident about: