Welcome to Differentiation: Mastering the Rate of Change!
Hello future mathematician! You’ve reached Unit P4, and differentiation is about to get much more powerful. In previous units (P3), you learned the basics of finding gradients; now, we are going to tackle functions that are nested, multiplied, divided, or defined in terms of other variables.
Why is this important? Differentiation is the backbone of calculus. It allows engineers to model the fastest flow of water, economists to find maximum profits, and physicists to determine instantaneous velocity and acceleration. Mastering these advanced rules will unlock your ability to solve complex, real-world problems.
Don’t worry if some parts look tricky at first. We will break down every rule into simple, manageable steps, complete with memory tricks and examples!
1. Revisiting the Foundation (A Quick P3 Prerequisite)
Before jumping into P4, let's quickly remind ourselves of the core concept and notation:
- The derivative \( \frac{dy}{dx} \) represents the instantaneous rate of change or the gradient of the tangent to the curve \(y = f(x)\).
- Power Rule Refresher: If \( y = ax^n \), then \( \frac{dy}{dx} = anx^{n-1} \).
We are now building a toolkit to differentiate functions that don't look like simple powers!
2. The Chain Rule: Differentiating a Function of a Function
The Concept: The Onion Analogy
The Chain Rule is used when one function is 'nested' inside another. Think of it like peeling an onion:
You must differentiate the outer layer first, keeping the inner layer intact, and then multiply by the derivative of the inner layer.
The Chain Rule Formula (Formal Definition)
If \( y \) is a function of \( u \), and \( u \) is a function of \( x \), then:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
The Simplified Trick (The Rule in Practice)
If \( y = f(g(x)) \):
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]
Step 1: Differentiate the Outside function, leaving the Inside function exactly as it is.
Step 2: Multiply the result by the derivative of the Inside function.
Example: Differentiate \( y = (3x^2 + 5)^4 \).
Here, the outside function is \( (\cdot)^4 \) and the inside is \( 3x^2 + 5 \).
- Outside: Bring the 4 down and reduce the power: \( 4(3x^2 + 5)^3 \).
- Inside: Differentiate the inside: \( \frac{d}{dx}(3x^2 + 5) = 6x \).
- Multiply: \( \frac{dy}{dx} = 4(3x^2 + 5)^3 \cdot (6x) = 24x(3x^2 + 5)^3 \).
Quick Takeaway for the Chain Rule
The Chain Rule is essential for composite functions, especially when dealing with high powers, trigonometric functions like \( \sin(2x) \), or exponential functions like \( e^{x^2} \).
3. The Product Rule: Multiplying Two Functions
If you have two functions of \(x\) multiplied together (e.g., \( x^2 \sin x \)), you cannot simply differentiate them separately. You need the Product Rule.
The Product Rule Formula
If \( y = uv \), where \( u \) and \( v \) are functions of \( x \):
\[ \frac{dy}{dx} = v \frac{du}{dx} + u \frac{dv}{dx} \]
Memory Aid (The Mnemonic)
Think of it as Vee Dee U plus U Dee Vee.
Step-by-Step Process
Example: Differentiate \( y = x^3 \cos x \).
- Define u and v:
\( u = x^3 \)
\( v = \cos x \) - Find the derivatives:
\( \frac{du}{dx} = 3x^2 \)
\( \frac{dv}{dx} = -\sin x \) - Substitute into the formula \( v \frac{du}{dx} + u \frac{dv}{dx} \):
\( \frac{dy}{dx} = (\cos x)(3x^2) + (x^3)(-\sin x) \) - Simplify:
\( \frac{dy}{dx} = 3x^2 \cos x - x^3 \sin x \)
Common Mistake to Avoid: A common beginner error is trying to differentiate $u$ and $v$ separately and multiplying the derivatives. \( \frac{d}{dx}(uv) \neq \frac{du}{dx} \frac{dv}{dx} \). You must use the full Product Rule!
4. The Quotient Rule: Dividing Two Functions
When one function of \(x\) is divided by another (e.g., \( \frac{\sin x}{x^2} \)), we use the Quotient Rule.
The Quotient Rule Formula
If \( y = \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \):
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Memory Aid (The Song/Mnemonic)
This is arguably the most famous calculus mnemonic. If you call \(u\) "High" and \(v\) "Low":
"Low Dee High minus High Dee Low, over Low Squared."
Example: Differentiate \( y = \frac{\ln x}{x} \).
- Define u and v:
\( u = \ln x \) (High)
\( v = x \) (Low) - Find the derivatives:
\( \frac{du}{dx} = \frac{1}{x} \)
\( \frac{dv}{dx} = 1 \) - Substitute into the formula \( \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \):
Numerator: \( (x)(\frac{1}{x}) - (\ln x)(1) = 1 - \ln x \) Denominator: \( (x)^2 = x^2 \) - Final Result:
\( \frac{dy}{dx} = \frac{1 - \ln x}{x^2} \)
Key Point: The order matters here! If you mix up \(u\) and \(v\) in the numerator, your sign will be wrong. Always start with Low Dee High!
5. Differentiating Exponential and Logarithmic Functions
These functions are incredibly common in nature (growth, decay, and financial modelling), so knowing their derivatives is crucial.
5.1 Exponential Functions (The Magic Function)
The derivative of \( e^x \) is uniquely simple:
\[ \frac{d}{dx} (e^x) = e^x \]
Did you know? \( e^x \) is the only function (other than multiples of itself) that is equal to its own derivative. This is why it’s called the natural exponential function.
Combining with the Chain Rule
If the power is not just \( x \), you MUST use the Chain Rule.
If \( y = e^{f(x)} \), then \( \frac{dy}{dx} = f'(x) e^{f(x)} \).
Example: Differentiate \( y = e^{3x^2 - 1} \).
The derivative of the power \( 3x^2 - 1 \) is \( 6x \).
\( \frac{dy}{dx} = (6x) e^{3x^2 - 1} \).
5.2 Natural Logarithmic Functions
The derivative of the natural logarithm (\(\ln x\)) is also simple:
\[ \frac{d}{dx} (\ln x) = \frac{1}{x} \]
Combining with the Chain Rule
If the input is a function of \( x \):
\[ \frac{d}{dx} (\ln(f(x))) = \frac{f'(x)}{f(x)} \]
Example: Differentiate \( y = \ln(5x+2) \).
The derivative of the inside \( (5x+2) \) is 5.
\( \frac{dy}{dx} = \frac{5}{5x+2} \).
A Note on Implicit Functions (Related to Logs)
Sometimes you need to differentiate \( \ln(kx) \). Remember that due to logarithm laws, \( \ln(kx) = \ln k + \ln x \). Since \(\ln k\) is a constant, its derivative is zero. Thus, \( \frac{d}{dx} (\ln(kx)) = \frac{1}{x} \).
6. Differentiating Trigonometric Functions
While you learned \(\sin x\) and \(\cos x\) derivatives in P3, P4 requires you to master the derivative of \(\tan x\) and combining all trig derivatives with the Chain Rule.
Core P4 Trigonometric Derivative
\[ \frac{d}{dx} (\tan x) = \sec^2 x \]
(This result can be derived using the Quotient Rule on \( \tan x = \frac{\sin x}{\cos x} \).)
The Cyclic Derivatives (Review)
Remember the P3 cycle:
- \( \sin x \rightarrow \cos x \)
- \( \cos x \rightarrow -\sin x \)
- \( -\sin x \rightarrow -\cos x \)
- \( -\cos x \rightarrow \sin x \) (back to the start)
The Chain Rule with Trig Functions
Whenever the angle is anything other than a simple \( x \), use the Chain Rule.
Example 1: Differentiate \( y = \cos(4x) \).
Outside derivative (\(\cos \rightarrow -\sin\)): \( -\sin(4x) \).
Inside derivative (of \( 4x \)): \( 4 \).
\( \frac{dy}{dx} = -4\sin(4x) \).
Example 2: Differentiate \( y = \tan^3 x \).
(Rewrite as \( y = (\tan x)^3 \). Inside is \(\tan x\). Outside is \( (\cdot)^3 \).)
Outside: \( 3(\tan x)^2 \).
Inside derivative (\(\frac{d}{dx} \tan x \)): \( \sec^2 x \).
\( \frac{dy}{dx} = 3\tan^2 x \sec^2 x \).
Quick Review: Key Derivatives
| Function \(y\) | Derivative \(\frac{dy}{dx}\) |
|---|---|
| \(e^x\) | \(e^x\) |
| \(\ln x\) | \(1/x\) |
| \(\sin x\) | \(\cos x\) |
| \(\cos x\) | \(-\sin x\) |
| \(\tan x\) | \(\sec^2 x\) |
7. Implicit Differentiation
So far, all functions have been explicitly defined, meaning \(y\) is isolated on one side (e.g., \( y = x^2 + 3 \)). Implicit differentiation is used when \(x\) and \(y\) are mixed together in an equation and it’s difficult or impossible to isolate \(y\). (E.g., \( x^2 + y^2 = 25 \) or \( xy + e^y = 1 \)).
The Core Rule: Differentiating \(y\) terms
When you differentiate a term involving \(y\) with respect to \(x\), you must apply the Chain Rule. Since \(y\) is assumed to be a function of \(x\), you differentiate the term as usual, and then multiply by \(\frac{dy}{dx}\).
Think of \(\frac{dy}{dx}\) as the "Chain Rule tax" you pay every time you differentiate a \(y\) term.
Example 1: Differentiate \( y^3 \) with respect to \( x \).
Differentiate the outside (\( y^3 \rightarrow 3y^2 \)).
Multiply by the tax: \( \frac{d}{dx} (y^3) = 3y^2 \frac{dy}{dx} \).
Step-by-Step Implicit Differentiation
Example: Find \(\frac{dy}{dx}\) for the circle \( x^2 + y^2 = 25 \).
- Differentiate Term by Term (with respect to \(x\)):
\( \frac{d}{dx} (x^2) = 2x \)
\( \frac{d}{dx} (y^2) = 2y \frac{dy}{dx} \) (Remember the tax!)
\( \frac{d}{dx} (25) = 0 \) (Constant derivative) - Write the Differentiated Equation:
\( 2x + 2y \frac{dy}{dx} = 0 \) - Isolate \(\frac{dy}{dx}\):
\( 2y \frac{dy}{dx} = -2x \)
\( \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y} \)
Implicit Differentiation and the Product Rule
If you have a mixed term like \( xy \), you must treat it as a product of two functions of \(x\): \( u=x \) and \( v=y \).
\[ \frac{d}{dx} (xy) = v \frac{du}{dx} + u \frac{dv}{dx} \]
\[ \frac{d}{dx} (xy) = (y)(1) + (x)\left(\frac{dy}{dx}\right) = y + x \frac{dy}{dx} \]
Summary: Implicit differentiation allows you to find gradients even when \(x\) and \(y\) are inseparable. The key is to remember to attach \( \frac{dy}{dx} \) to every term you differentiate that involves \(y\).
8. Parametric Differentiation
Sometimes, the coordinates \(x\) and \(y\) are both defined in terms of a third variable, often time \(t\) or an angle \(\theta\). This third variable is called the parameter.
Example: \( x = 2t^2 \) and \( y = 4t \).
We want to find the gradient \( \frac{dy}{dx} \), but we only have \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
The Parametric Differentiation Formula
Using the Chain Rule, we can write:
\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \]
Since \( \frac{dt}{dx} = \frac{1}{dx/dt} \), the practical formula becomes:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Step-by-Step Parametric Differentiation
Example: Find \( \frac{dy}{dx} \) for the curve defined by \( x = \cos \theta \) and \( y = \sin 2\theta \).
- Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):
\( \frac{dx}{d\theta} = -\sin \theta \)
\( \frac{dy}{d\theta} = 2\cos 2\theta \) (Remember the Chain Rule!) - Apply the formula:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\cos 2\theta}{-\sin \theta} \] - Simplify (Optional, but often required):
Using the identity \(\cos 2\theta = 1 - 2\sin^2 \theta\):
\[ \frac{dy}{dx} = \frac{2(1 - 2\sin^2 \theta)}{-\sin \theta} = \frac{-2 + 4\sin^2 \theta}{\sin \theta} \]
Key Takeaway: Parametric differentiation is simpler than it looks. It is just a division of two derivatives you already know how to find! Just ensure you correctly apply the Chain Rule when differentiating \(x\) and \(y\) with respect to the parameter.
9. Applications of Differentiation (P4 Context)
9.1 Tangents and Normals
These applications remain the same as in P3, but now you use the new rules (Chain, Product, Quotient, Implicit, Parametric) to find the gradient \(\frac{dy}{dx}\).
- Gradient of the Tangent: \( m_t = \frac{dy}{dx} \) evaluated at a specific point.
- Gradient of the Normal: \( m_n = -\frac{1}{m_t} \) (the negative reciprocal).
- Use the equation \( y - y_1 = m(x - x_1) \) to find the line equation.
Crucial Step in Parametric/Implicit Questions: If you are given a point \((x_1, y_1)\), for implicit curves you plug both \(x_1\) and \(y_1\) into the derivative formula. For parametric curves, you must first find the value of the parameter (\(t\) or \(\theta\)) that corresponds to the given point.
Example: If \( x = t^2 \) and you need the tangent at \( x=4 \), first solve \( 4 = t^2 \) to find \( t=\pm 2 \). You must calculate \(\frac{dy}{dx}\) for both values of \(t\) if applicable!
9.2 Rates of Change
This application relies entirely on using the Chain Rule to link different variables together. If a problem asks for the rate of change of volume \(V\) with respect to time \(t\), i.e., \( \frac{dV}{dt} \), and you know the rate of change of radius \(r\) with respect to time, \( \frac{dr}{dt} \), you use:
\[ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \]
Imagine these fractions like a chain of train cars—the intermediate variable (\(dr\)) links the start (\(dV\)) to the end (\(dt\)).
Always define your rates clearly before attempting the question!
"The radius is increasing at 2 cm/s" means \( \frac{dr}{dt} = 2 \).
Final Encouragement: You now possess the most essential tools in differential calculus. Practice applying the rules systematically, and remember the mnemonics. You’ve got this!