Implicit, parametric and logarithmic differentiation

IB Mathematics Analysis and Approaches HL· AA HL 5.4 Differentiation techniques· 15 min read

1. Implicit Differentiation★★★☆☆⏱ 5 min

🚫 No Calculator

📘 Definition

Implicit Differentiation

The process of differentiating both sides of an implicit relation with respect to $x$ , applying the chain rule to all terms containing $y$ .

Example:

Used for relations like $x^{2} + y^{3} = x y$ that are hard to rearrange to $y = f (x)$ .

By the chain rule, any function of $y$ differentiated with respect to $x$ gives the identity:

\frac{d}{dx}\left[f(y)\right] = f'(y) \cdot \frac{dy}{dx}

📐 Worked Example

Find $\frac{d y}{d x}$ for $x^{2} + y^{2} = 25$ , then calculate the gradient at $(3, 4)$ .

1
Differentiate each term on both sides with respect to $x$ :
2
$2x + 2y \cdot \frac{dy}{dx} = 0$
3
Rearrange to isolate $\frac{d y}{d x}$ :
4
$2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$
5
Substitute the point $(3, 4)$ to get the gradient:
6
$\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}$

Exam tip:

Always remember the chain rule factor for terms containing $y$ .

2. Parametric Differentiation★★★★☆⏱ 5 min

🚫 No Calculator

For parametric functions $x = x (t)$ and $y = y (t)$ , we rearrange the chain rule to get the derivative of $y$ with respect to $x$ :

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \frac{dx}{dt} \neq 0

For the second derivative, we apply the chain rule again, since we are differentiating with respect to $x$ , not $t$ :

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

📐 Worked Example

Given $x = 2 t^{3} + 1$ , $y = 4 t^{2} - t$ , find $\frac{d y}{d x}$ in terms of $t$ , then find the gradient at $t = 1$ .

1
Differentiate $x$ and $y$ separately with respect to $t$ :
2
$\frac{dx}{dt} = 6t^2, \quad \frac{dy}{dt} = 8t - 1$
3
Apply the parametric derivative formula:
4
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{8t - 1}{6t^2}$
5
Substitute $t = 1$ to get the gradient:
6
$\frac{dy}{dx}\bigg|_{t=1} = \frac{8(1) - 1}{6(1)^2} = \frac{7}{6}$

3. Logarithmic Differentiation★★★★☆⏱ 5 min

🚫 No Calculator

Logarithmic differentiation is the go-to technique for two types of functions that are hard to differentiate with basic rules: 1) Functions of the form $y = f (x)^{g (x)}$ (variable base and variable exponent), and 2) Complicated products, quotients, or powers of multiple functions.

The method follows four core steps: 1) Take the natural logarithm of both sides, 2) Expand using logarithm laws, 3) Differentiate implicitly with respect to $x$ , 4) Multiply by $y$ to isolate $\frac{d y}{d x}$ .

📐 Worked Example

Differentiate $y = \frac{x ^{2} x + 1}{( x + 2 ) ^{3}}$ with respect to $x$ .

1
Take natural logarithm of both sides, then expand using log rules:
2
$\ln y = 2\ln x + \frac{1}{2}\ln(x+1) - 3\ln(x+2)$
3
Differentiate both sides implicitly with respect to $x$ :
4
$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{2}{x} + \frac{1}{2(x+1)} - \frac{3}{x+2}$
5
Multiply by $y$ and substitute the original expression for $y$ :
6
$\frac{dy}{dx} = \frac{x^2 \sqrt{x+1}}{(x+2)^3} \left( \frac{2}{x} + \frac{1}{2(x+1)} - \frac{3}{x+2} \right)$

4. Common Pitfalls

Wrong move:

Forgetting to multiply by $\frac{d y}{d x}$ after differentiating a $y$ -term in implicit differentiation.

Why:

You are differentiating with respect to $x$ , not $y$ , so the chain rule requires the extra factor.

Correct move:

Add $\frac{d y}{d x}$ as a factor every time you differentiate a function of $y$ .

Wrong move:

Calculating the second derivative of a parametric function by differentiating $\frac{d y}{d x}$ directly with respect to $t$ .

Why:

$\frac{d ^{2} y}{d x ^{2}}$ is the derivative with respect to $x$ , not $t$ , so the chain rule is required.

Correct move:

Use the identity $\frac{d ^{2} y}{d x ^{2}} = \frac{1}{d x / d t} \cdot \frac{d}{d t} (\frac{d y}{d x})$ .

Wrong move:

Swapping numerator and denominator in the parametric derivative formula: $\frac{d y}{d x} = \frac{d x / d t}{d y / d t}$ .

Why:

This is a common order mix-up when memorizing the formula.

Correct move:

Remember: dy over dx equals (dy over dt) divided by (dx over dt).

Wrong move:

Forgetting to multiply by the original $y$ after implicit differentiation in logarithmic differentiation.

Why:

After expanding and differentiating, you are left with $\frac{1}{y} \frac{d y}{d x} = ...$ , so the $y$ factor is required.

Correct move:

Always end by multiplying through by $y$ and substitute back the original expression for $y$ .

Wrong move:

Using logarithmic differentiation for simple polynomials like $y = x^{5}$ when power rule works.

Why:

It is unnecessary and wastes exam time, increasing your chance of making an arithmetic error.

Correct move:

Only use logarithmic differentiation for complex products/quotients or functions of the form $f (x)^{g (x)}$ .

5. Quick Reference Cheatsheet

Technique	Core Formula	Use Case
Implicit Differentiation	Differentiate term-by-term: $\frac{d}{d x} [f (y)] = f^{'} (y) \frac{d y}{d x}$ , then rearrange	For relations that can't be written as $y = f (x)$
Parametric Differentiation	$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ , $\frac{d ^{2} y}{d x ^{2}} = \frac{d / d t ( d y / d x )}{d x / d t}$	When $x, y$ are both functions of a parameter $t$
Logarithmic Differentiation	$ln y = ln f (x) ⟹ \frac{1}{y} \frac{d y}{d x} = \frac{f ^{'} ( x )}{f ( x )} ⟹ \frac{d y}{d x} = y \cdot \frac{f ^{'} ( x )}{f ( x )}$	For $y = f (x)^{g (x)}$ , complicated products/quotients

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · Paper 1
Find gradient of implicit curve
2022 · Paper 2
Differentiate parametric function
2023 · Paper 1
Log differentiation of complex product

Going deeper

VideoImplicit differentiation worked examplesExtra practice for core concepts

What's Next

Mastering these three differentiation techniques is critical for all remaining calculus topics in IB AA HL. You will next apply these methods to find tangents, normals, and stationary points of implicit and parametric curves, and solve popular exam-related rates problems. These techniques also form the foundation for integration of parametric functions, and calculating arc lengths, surface areas, and volumes of revolution in later units. Most IB exam questions combine these techniques with other differentiation topics, so consistent practice will help you recognize when to use each method efficiently.