Library Pure Mathematics 4 WMA14 Implicit Differentiation
A2 Level · Pure Mathematics 4 WMA14

Implicit Differentiation

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Implicit Differentiation

Finding the slope when equations won't let you isolate y

What You Need to Know

The Problem Some equations mix x and y so thoroughly you can't rearrange them into y = f(x)
The Solution Differentiate both sides with respect to x, treating y as a function of x
Key Tool The chain rule: whenever you differentiate a y-term, multiply by dy/dx
The Rearrangement After differentiating, collect all dy/dx terms on one side and solve

1. Why Implicit Differentiation?

In most calculus problems, you're given equations in the form y = f(x)—these are called explicit equations. You just differentiate the right side and you're done.

But some equations are messy. Take:

sin y = 3x²e^(-4y)

Try to rearrange this to get y alone. You can't. The x and y are completely tangled together. This is an implicit equation—y is hidden implicitly in the relationship, rather than explicitly isolated.

But we still need to find dy/dx. That's where implicit differentiation comes in: we differentiate both sides of the equation with respect to x, and the chain rule does the heavy lifting.

Key insight: You don't need to solve for y. You don't even need y explicitly. You just need an expression for dy/dx in terms of both x and y, and that's useful enough.

2. The Chain Rule for y-Functions

The foundation of implicit differentiation is this: whenever you differentiate a function of y with respect to x, you must multiply by dy/dx.

Let's say f(y) is some function of y, like sin(y) or e^y or y³.

Chain Rule for y-Functions
d/dx[f(y)] = f'(y) · dy/dx

Why?

Because y itself depends on x. By the chain rule:

d/dx[f(y)] = df/dy · dy/dx

When you differentiate a y-function, you get the derivative with respect to y, then you multiply by dy/dx (the rate of change of y with respect to x).

Example: sin(y)

The derivative of sin(y) with respect to y is cos(y). So:

d/dx[sin y] = cos y · dy/dx
Remember: Every y-term gets a dy/dx tag attached when you differentiate. X-terms don't—they're independent variables.
What is d/dx[e^(2y)]?

3. Product Rule + Chain Rule = Power

When a term involves both x and y multiplied together, the product rule and chain rule work together beautifully.

Product Rule with y-Functions
d/dx[f(x)·g(y)] = f'(x)·g(y) + f(x)·g'(y)·dy/dx

In words: Differentiate the first function (x-function) and leave the second alone. Then leave the first and differentiate the second (remembering the dy/dx). Add them together.

Example: 3x² · e^(-4y)

Here f(x) = 3x² and g(y) = e^(-4y).

d/dx[3x² · e^(-4y)] = 6x · e^(-4y) + 3x² · (-4e^(-4y)) · dy/dx = 6x·e^(-4y) - 12x²·e^(-4y) · dy/dx
Common pattern: When you have a term like xy or x²y or x·sin(y), you always use the product rule, and the y-part gets a dy/dx attached.
Find d/dx[2x · y]

4. Four Essential Formulas (Memorise These)

These four formulas capture the most common patterns you'll see. Learn them once, use them forever.

Formula 1: Chain Rule for y-Functions

When you see f(y)
d/dx[y^n] = n·y^(n-1) · dy/dx

Special case (power rule): If you have y³ or y^(1/2) or any power of y, just apply the power rule and attach dy/dx.

Examples:

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Also in the full note
  • 5. Step-by-Step Method
  • 6. Critical Mistakes to Avoid
  • 7. What the Examiners Look For
  • What to Memorise
  • Concepts Checklist
  • Final Practice
  • Formula 2: Product Rule (x and y)
  • Formula 3: Power Rule with y
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