Welcome to Implicit Differentiation!

Up until now, you have probably spent most of your time differentiating functions where y is all by itself on one side, like \(y = 3x^2 + 5\). This is called an explicit function because it explicitly tells you what y is.

But what happens when x and y are all mixed up together, like in the equation of a circle: \(x^2 + y^2 = 25\)? Trying to get y on its own can be messy and sometimes impossible! Don't worry if this seems tricky at first; implicit differentiation is just a clever way to find the gradient (\(\frac{dy}{dx}\)) without needing to untangle the equation first.

Section Summary: We use implicit differentiation when y is not the subject of the formula.

1. Explicit vs. Implicit: What’s the difference?

Think of an explicit function like a pre-made sandwich: you can see exactly what's inside (\(y = \text{ingredients}\)).
An implicit relation is like a burrito: the ingredients (x and y) are all wrapped up together (\(x^2 + y^2 = 25\)).

  • Explicit: \(y = 2x + 3\). To find the gradient, we just differentiate as usual.
  • Implicit: \(x^2 + y^2 = 10\). To find the gradient, we differentiate every term just as it sits.

Did you know? Many beautiful curves in nature and engineering, like the orbits of planets or the shape of a cooling tower, are naturally described by implicit equations!

2. The "Golden Rule" of Implicit Differentiation

The secret ingredient here is the Chain Rule. Because y is actually a function of x, every time you differentiate a term containing y, you must "tag" it with a \(\frac{dy}{dx}\).

Quick Review: The Chain Rule Logic

If you have \(y^2\) and you want to differentiate it with respect to \(x\):
1. Differentiate it with respect to y (which gives you \(2y\)).
2. Multiply by \(\frac{dy}{dx}\).
So, the derivative of \(y^2\) is \(2y \frac{dy}{dx}\).

Memory Aid: "Differentiate and Tag"
Differentiate the y term normally, then "tag" it with a \(\frac{dy}{dx}\). If you differentiate an x term, you don't need a tag!

Key Takeaway: \(\frac{d}{dx}(f(y)) = f'(y) \frac{dy}{dx}\).

3. Step-by-Step Guide to Success

Let's find \(\frac{dy}{dx}\) for the curve \(x^2 + y^3 = 5x\).

Step 1: Differentiate everything with respect to x.
The derivative of \(x^2\) is \(2x\).
The derivative of \(y^3\) is \(3y^2 \frac{dy}{dx}\) (Remember the tag!).
The derivative of \(5x\) is \(5\).
Our equation now looks like: \(2x + 3y^2 \frac{dy}{dx} = 5\)

Step 2: Get all terms with \(\frac{dy}{dx}\) on one side.
\(3y^2 \frac{dy}{dx} = 5 - 2x\)

Step 3: Solve for \(\frac{dy}{dx}\).
Divide both sides by \(3y^2\):
\(\frac{dy}{dx} = \frac{5 - 2x}{3y^2}\)

Quick Review Box:
1. Differentiate everything (\(x\) terms normally, \(y\) terms with a tag).
2. Move \(\frac{dy}{dx}\) terms to the left.
3. Move everything else to the right.
4. Factorise and divide.

4. The Product Rule Trap

This is where most students lose marks! If you see a term like \(xy\) or \(x^2y\), you must use the Product Rule because you are multiplying two functions (\(x\) and \(y\)).

Example: Differentiate \(xy\).
Let \(u = x\) and \(v = y\).
\(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{dy}{dx}\).
Using \(u\frac{dv}{dx} + v\frac{du}{dx}\), we get: \(x\frac{dy}{dx} + y(1)\).
So, \(\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\).

Common Mistake to Avoid: Writing the derivative of \(xy\) as just \(\frac{dy}{dx}\). Always check if x and y are multiplying each other!

5. Real-World Example: The Circle

Let's find the gradient of the circle \(x^2 + y^2 = 25\) at the point \((3, 4)\).

1. Differentiate: \(2x + 2y\frac{dy}{dx} = 0\) (The derivative of a constant like 25 is always 0!).
2. Rearrange: \(2y\frac{dy}{dx} = -2x\).
3. Isolate: \(\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\).
4. Substitute the point \((3, 4)\): Gradient \(= -\frac{3}{4}\).

Key Takeaway: For implicit differentiation, your answer for \(\frac{dy}{dx}\) will usually contain both x and y. This is perfectly normal!

Summary Checklist

  • Can I identify an implicit equation? (Mixed x and y).
  • Do I remember to add \(\frac{dy}{dx}\) every time I differentiate a y term?
  • Am I watching out for the Product Rule (e.g., \(xy\))?
  • Do I remember that the derivative of a constant (like 10 or 100) is 0?

Final Tip: Keep your work neat! It’s easy to lose a \(\frac{dy}{dx}\) in the middle of a long equation. Using brackets can help keep your "tagged" terms clear.