# Implicit, parametric and logarithmic differentiation

> IB Mathematics Analysis and Approaches HL · Unit 5: Calculus
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-implicit-parametric-and-logarithmic-differentiation/

This module covers three advanced differentiation techniques for functions that cannot be written in explicit $y=f(x)$ form. You will learn to extend the chain rule to solve for derivatives of implicit, parametric, and complex algebraic functions common in IB exams.

**Prerequisites:** [Basic differentiation rules (product, quotient, chain)](https://www.owlsprep.com/study/ib-math-aa-hl-u5-basic-differentiation-rules/); [Differentiation of exponential and logarithmic functions](https://www.owlsprep.com/study/ib-math-aa-hl-u5-exponential-log-differentiation/)

## Learning objectives

- Differentiate implicit functions using the chain rule
- Calculate first and second derivatives of parametric functions
- Apply logarithmic differentiation to complex algebraic functions
- Evaluate gradients at specified points for all three techniques

## Implicit Differentiation

**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 $x^2 + y^3 = xy$ 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 $\frac{dy}{dx}$ 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 $\frac{dy}{dx}$:
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 $y$.

*Calculator:* forbidden

## Parametric Differentiation

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 $x = 2t^3 + 1$, $y = 4t^2 - t$, find $\frac{dy}{dx}$ 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. $$\frac{dy}{dx}\bigg|_{t=1} = \frac{8(1) - 1}{6(1)^2} = \frac{7}{6}$$

> **warning**
>
> Never just differentiate $\frac{dy}{dx}$ with respect to $t$ for the second derivative — you must divide by $\frac{dx}{dt}$.

*Calculator:* forbidden

## Logarithmic Differentiation

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 $\frac{dy}{dx}$.

**Worked example:** Differentiate $y = \frac{x^2 \sqrt{x+1}}{(x+2)^3}$ 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. $$\frac{dy}{dx} = \frac{x^2 \sqrt{x+1}}{(x+2)^3} \left( \frac{2}{x} + \frac{1}{2(x+1)} - \frac{3}{x+2} \right)$$

> **tip**
>
> This method is much faster than repeated product/quotient rule for this type of problem, so always use it for variable exponents.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Forgetting to multiply by $\frac{dy}{dx}$ after differentiating a $y$-term in implicit differentiation.
  - Why it fails: You are differentiating with respect to $x$, not $y$, so the chain rule requires the extra factor.
  - Correct: Add $\frac{dy}{dx}$ as a factor every time you differentiate a function of $y$.
- **Wrong:** Calculating the second derivative of a parametric function by differentiating $\frac{dy}{dx}$ directly with respect to $t$.
  - Why it fails: $\frac{d^2y}{dx^2}$ is the derivative with respect to $x$, not $t$, so the chain rule is required.
  - Correct: Use the identity $\frac{d^2y}{dx^2} = \frac{1}{dx/dt} \cdot \frac{d}{dt}\left(\frac{dy}{dx}\right)$.
- **Wrong:** Swapping numerator and denominator in the parametric derivative formula: $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$.
  - Why it fails: This is a common order mix-up when memorizing the formula.
  - Correct: Remember: dy over dx equals (dy over dt) divided by (dx over dt).
- **Wrong:** Forgetting to multiply by the original $y$ after implicit differentiation in logarithmic differentiation.
  - Why it fails: After expanding and differentiating, you are left with $\frac{1}{y} \frac{dy}{dx} = ...$, so the $y$ factor is required.
  - Correct: Always end by multiplying through by $y$ and substitute back the original expression for $y$.
- **Wrong:** Using logarithmic differentiation for simple polynomials like $y = x^5$ when power rule works.
  - Why it fails: It is unnecessary and wastes exam time, increasing your chance of making an arithmetic error.
  - Correct: Only use logarithmic differentiation for complex products/quotients or functions of the form $f(x)^{g(x)}$.

## Cheatsheet

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

## 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.

- [Higher order derivatives](https://www.owlsprep.com/study/ib-math-aa-hl-u5-higher-order-derivatives/)
- [Applications of Differentiation](https://www.owlsprep.com/study/ib-math-aa-hl-u5-applications-of-differentiation/)
- [Indefinite and definite integration](https://www.owlsprep.com/study/ib-math-aa-hl-u5-indefinite-and-definite-integration/)

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