Mastering Implicit Differentiation: Your A-Level Maths Revision Guide

What the examiner is testing

The core skill assessed is differentiating expressions where \(y\) is not explicitly defined as a function of \(x\), specifically applying the chain rule correctly to terms involving \(y\). Marks are awarded for each correctly differentiated term and for the final rearrangement to isolate \(\frac{dy}{dx}\).

The method

1. Differentiate every term with respect to \(x\).
2. When differentiating a term involving \(y\), treat \(y\) as an inner function of \(x\), so apply the chain rule: differentiate with respect to \(y\) and then multiply by \(\frac{dy}{dx}\).
3. Collect all terms containing \(\frac{dy}{dx}\) on one side of the equation and all other terms on the opposite side.
4. Factorise out \(\frac{dy}{dx}\).
5. Divide by the remaining expression to isolate \(\frac{dy}{dx}\).

Worked example

Find \(\frac{dy}{dx}\) for the equation \(x^2 + y^3 = 5\).

$$ x^2 + y^3 = 5 $$

Differentiate each term with respect to \(x\):

$$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(5) $$

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

Collect terms with \(\frac{dy}{dx}\) on one side:

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

Isolate \(\frac{dy}{dx}\):

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

Sanity check: If \(y\) were explicitly \(y = (5-x^2)^{1/3}\), then \(\frac{dy}{dx} = \frac{1}{3}(5-x^2)^{-2/3}(-2x) = -\frac{2x}{3(5-x^2)^{2/3}}\). Since \(y^3 = 5-x^2\), then \(y^2 = (5-x^2)^{2/3}\), matching our result.

Worked example: a harder one

Find \(\frac{dy}{dx}\) for the equation \(x \sin y = y e^x\).

$$ x \sin y = y e^x $$

The obvious first move is to differentiate each side directly. However, both sides involve products, so we must apply the product rule.

Differentiate \(x \sin y\) using the product rule \((uv)' = u'v + uv'\), where \(u=x\) and \(v=\sin y\).
\(\frac{d}{dx}(x) = 1\)
\(\frac{d}{dx}(\sin y) = \cos y \frac{dy}{dx}\) (using the chain rule)
So, \(\frac{d}{dx}(x \sin y) = (1)(\sin y) + (x)(\cos y \frac{dy}{dx}) = \sin y + x \cos y \frac{dy}{dx}\).

Differentiate \(y e^x\) using the product rule, where \(u=y\) and \(v=e^x\).
\(\frac{d}{dx}(y) = \frac{dy}{dx}\)
\(\frac{d}{dx}(e^x) = e^x\)
So, \(\frac{d}{dx}(y e^x) = (\frac{dy}{dx})(e^x) + (y)(e^x) = e^x \frac{dy}{dx} + y e^x\).

Equating the differentiated sides:

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

Collect terms with \(\frac{dy}{dx}\) on one side (e.g., left) and other terms on the other side:

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

Factorise out \(\frac{dy}{dx}\):

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

Isolate \(\frac{dy}{dx}\):

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

Practice

1. Find \(\frac{dy}{dx}\) for \(x^3 + y^3 = 6xy\).
2. Find \(\frac{dy}{dx}\) for \(e^y + \cos x = xy\).
3. The curve \(C\) has equation \(x^2 + 4xy - 3y^2 + 2x = 10\). Find an expression for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).
4. A curve is defined by the equation \(x^2 - 3xy + 4y^2 = 13\). Find the equation of the normal to the curve at the point \((2, 3)\).

Answers:
1. \(\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}\) or \(\frac{2y - x^2}{y^2 - 2x}\)
2. \(\frac{dy}{dx} = \frac{y + \sin x}{e^y - x}\)
3. \(\frac{dy}{dx} = \frac{2x + 4y + 2}{6y - 4x}\) or \(\frac{x + 2y + 1}{3y - 2x}\)
4. Working for question 4:
Differentiate \(x^2 - 3xy + 4y^2 = 13\) with respect to \(x\):
$$ 2x - (3y + 3x \frac{dy}{dx}) + 8y \frac{dy}{dx} = 0 $$
$$ 2x - 3y - 3x \frac{dy}{dx} + 8y \frac{dy}{dx} = 0 $$
Collect terms with \(\frac{dy}{dx}\):
$$ (8y - 3x) \frac{dy}{dx} = 3y - 2x $$
$$ \frac{dy}{dx} = \frac{3y - 2x}{8y - 3x} $$
At the point \((2, 3)\):
$$ \frac{dy}{dx} = \frac{3(3) - 2(2)}{8(3) - 3(2)} = \frac{9 - 4}{24 - 6} = \frac{5}{18} $$
The gradient of the tangent at \((2, 3)\) is \(\frac{5}{18}\).
The gradient of the normal is \(-\frac{1}{\text{gradient of tangent}} = -\frac{18}{5}\).
Equation of the normal using \(y - y_1 = m(x - x_1)\):
$$ y - 3 = -\frac{18}{5}(x - 2) $$
$$ 5(y - 3) = -18(x - 2) $$
$$ 5y - 15 = -18x + 36 $$
$$ 18x + 5y - 51 = 0 $$

The three mistakes that lose marks

1. Forgetting the \(\frac{dy}{dx}\) term: When differentiating a term involving \(y\), students often forget to multiply by \(\frac{dy}{dx}\).
   • Wrong answer produced: For \(y^2\), differentiating to \(2y\) instead of \(2y \frac{dy}{dx}\).
2. Incorrect application of the product/quotient rule: If a term contains both \(x\) and \(y\) (e.g., \(xy\) or \(x^2y\)), both the product rule and implicit differentiation must be applied.
   • Wrong answer produced: For \(xy\), differentiating to \(y\) instead of \(y + x \frac{dy}{dx}\).
3. Algebraic errors when rearranging: After differentiation, correctly isolating \(\frac{dy}{dx}\) requires careful collection of terms and factorisation.
   • Wrong answer produced: Forgetting to factor out \(\frac{dy}{dx}\) before dividing, or making sign errors when moving terms across the equals sign.

30-second recap

Implicit differentiation allows finding \(\frac{dy}{dx}\) when \(y\) isn't explicitly defined. Differentiate every term with respect to \(x\), remembering to multiply by \(\frac{dy}{dx}\) for any term involving \(y\). Then, rearrange algebraically to make \(\frac{dy}{dx}\) the subject.