Implicit Differentiation — Question Types

Type 1: First Derivative $\frac{dy}{dx}$ (3 marks)

Example: w20/22 Q5(a) — Find $\frac{dy}{dx}$ in terms of $x$ and $y$, given that $x^3 + y^3 = 3xy$.

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Differentiate implicitly:

$3x^2 + 3y^2\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$

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

$\frac{dy}{dx}(3y^2 - 3x) = 3y - 3x^2$

$\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}$

Marks scheme:

• M1 — Differentiate implicitly, including $\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$ and product rule for $3xy$
• A1 — Correct differentiated equation
• A1 — Correct expression for $\frac{dy}{dx}$

Type 2: Second Derivative $\frac{d^2y}{dx^2}$ (5 marks)

Example: w21/21 Q3 — Given $x^3 + y^3 = 3xy$, find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$.

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From part (a): $\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}$

Differentiate again:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{y - x^2}{y^2 - x}\right)$

Using quotient rule:

$\frac{d^2y}{dx^2} = \frac{(\frac{dy}{dx} - 2x)(y^2 - x) - (y - x^2)(2y\frac{dy}{dx} - 1)}{(y^2 - x)^2}$

Substitute $\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}$ and simplify using $x^3 + y^3 = 3xy$.

After simplification:

$\frac{d^2y}{dx^2} = \frac{2xy}{(y^2 - x)^3}$

Marks scheme:

• M1 — Attempt quotient rule on $\frac{dy}{dx}$
• M1 — Differentiate $y$ terms implicitly (including $\frac{dy}{dx}$)
• A1 — Correct expression before simplification
• A1 — Substitute expression for $\frac{dy}{dx}$
• A1 — Correct simplified $\frac{d^2y}{dx^2}$

Type 3: Values at Specific Points (8 marks total)

Example: s23/23 Q4 — The curve is defined by $x^2 + xy + y^2 = 7$. (a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$. [3] (b) Find the equation of the tangent at $(-1, 2)$. [2] (c) Find $\frac{d^2y}{dx^2}$ at $(-1, 2)$. [3]

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(a) Differentiate:

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

$\frac{dy}{dx}(x + 2y) = -(2x + y)$

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

M1 — Implicit differentiation with product rule for $xy$ A1 — Correct differentiated equation A1 — Correct $\frac{dy}{dx}$

(b) At $(-1, 2)$:

$\frac{dy}{dx} = -\frac{2(-1) + 2}{-1 + 2(2)} = -\frac{0}{3} = 0$

Tangent: $y - 2 = 0(x + 1)$ i.e. $y = 2$

M1 — Substitute point into $\frac{dy}{dx}$ A1 — Correct equation $y = 2$

(c) Differentiate $\frac{dy}{dx}$ expression:

$\frac{d^2y}{dx^2} = -\frac{(2 + \frac{dy}{dx})(x + 2y) - (2x + y)(1 + 2\frac{dy}{dx})}{(x + 2y)^2}$

At $(-1, 2)$ with $\frac{dy}{dx} = 0$:

$\frac{d^2y}{dx^2} = -\frac{(2)(3) - (0)(1)}{9} = -\frac{6}{9} = -\frac{2}{3}$

M1 — Attempt quotient rule A1 — Correct substitution of $\frac{dy}{dx}=0$ and values A1 — Correct $-\frac{2}{3}$

Example: s24/23 Q3 — Given $x\sin y + ye^x = 0$. (a) Find $\frac{dy}{dx}$ at $(0, \pi)$. [4] (b) Find $\frac{d^2y}{dx^2}$ at $(0, \pi)$. [4]

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(a) Differentiate:

$\sin y + x\cos y\frac{dy}{dx} + \frac{dy}{dx}e^x + ye^x = 0$

$\frac{dy}{dx}(x\cos y + e^x) = -\sin y - ye^x$

$\frac{dy}{dx} = -\frac{\sin y + ye^x}{x\cos y + e^x}$

At $(0, \pi)$: $\frac{dy}{dx} = -\frac{0 + \pi \cdot 1}{0\cdot(-1) + 1} = -\pi$

M1 — Product rule on $x\sin y$ and $ye^x$ A1 — Correct differentiated equation M1 — Rearrange for $\frac{dy}{dx}$ A1 — Correct $-\pi$

(b) Differentiate $\frac{dy}{dx}$:

Numerator: $N = -(\sin y + ye^x)$, Denominator: $D = x\cos y + e^x$

At $(0, \pi)$ with $\frac{dy}{dx} = -\pi$:

$N' = -(\cos y\frac{dy}{dx} + \frac{dy}{dx}e^x + ye^x)$, $D' = \cos y - x\sin y\frac{dy}{dx} + e^x$

$\frac{d^2y}{dx^2} = \frac{N'D - ND'}{D^2}$ evaluated at $(0, \pi)$

$N' = -((-1)(-\pi) + (-\pi)(1) + \pi(1)) = -(\pi - \pi + \pi) = -\pi$

$D' = -1 - 0 + 1 = 0$

$\frac{d^2y}{dx^2} = \frac{(-\pi)(1) - (-\pi)(0)}{1} = -\pi$

M1 — Attempt to differentiate $\frac{dy}{dx}$ M1 — Correct $N'$, $D'$ with chain rule A1 — Correct values at $(0, \pi)$ A1 — Correct $-\pi$