Welcome to the World of Differentiation!
Hello future mathematician! This chapter, Differentiation, is arguably one of the most powerful tools in Pure Mathematics. Don't worry if the formulas look complicated initially; we will break them down piece by piece. By the end of this unit, you will be able to calculate the instantaneous rate of change for almost any complex function.
What will we learn? We move beyond the basic polynomial differentiation from P1/P2 and introduce new functions (like $e^x$ and $\ln x$) and sophisticated techniques (like the Product and Quotient Rules) necessary to tackle complex, real-world problems.
1. Essential P3 Derivatives: Exponential, Logarithmic, and Trig Functions
In P1 and P2, we mainly differentiated $x^n$. Now we introduce the key functions that define P3 mathematics.
1.1. Exponential and Logarithmic Functions
The number e (approximately 2.718) is special because the gradient of the function $y = e^x$ is always equal to the value of the function itself!
The Rules:
- If \( y = e^x \), then \( \frac{dy}{dx} = e^x \).
- If \( y = e^{kx} \), then \( \frac{dy}{dx} = ke^{kx} \). (We multiply by the constant in the power.)
- If \( y = \ln x \), then \( \frac{dy}{dx} = \frac{1}{x} \). (This only works for \( x > 0 \).)
Example 1.1: Simple P3 Differentiation
Differentiate \( y = 5e^{3x} - 2\ln x \):
\begin{align*} \frac{dy}{dx} &= 5(3e^{3x}) - 2\left(\frac{1}{x}\right) \\ &= 15e^{3x} - \frac{2}{x} \end{align*}
1.2. Derivatives of Trigonometric Functions (P3 Scope)
You should recall the differentiation rules for sine and cosine from earlier modules, but P3 requires you to know the rules for tangent and the three reciprocal functions: secant, cosecant, and cotangent.
The Core Six Rules:
- If \( y = \sin x \), then \( \frac{dy}{dx} = \cos x \).
- If \( y = \cos x \), then \( \frac{dy}{dx} = -\sin x \). (Note the minus sign!)
- If \( y = \tan x \), then \( \frac{dy}{dx} = \sec^2 x \).
- If \( y = \sec x \), then \( \frac{dy}{dx} = \sec x \tan x \).
- If \( y = \csc x \), then \( \frac{dy}{dx} = -\csc x \cot x \).
- If \( y = \cot x \), then \( \frac{dy}{dx} = -\csc^2 x \).
Memory Aid (Negative Signs): Notice that any trigonometric function that starts with a "c" ($\cos x$, $\csc x$, $\cot x$) has a derivative that starts with a negative sign.
Memorise the six trigonometric differentiation rules above! You must be able to recall them instantly.
2. The Chain Rule (Function of a Function)
What happens if you need to differentiate a function like \( y = \sin(x^2) \) or \( y = (3x+1)^5 \)? These are composite functions (a function inside another function). We use the Chain Rule.
Analogy: The Russian Doll
Imagine a Russian nesting doll. To open the whole thing, you must first open the outer doll, and then open the inner doll. The Chain Rule works the same way: differentiate the outer function, then multiply by the derivative of the inner function.
2.1. The Chain Rule Formula
If \( y \) is a function of \( u \), and \( u \) is a function of \( x \), then:
\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]
2.2. Step-by-Step Application
Goal: Differentiate \( y = e^{\sin x} \)
- Identify the Inner Function (\( u \)): This is the part inside the main function.
\( u = \sin x \)
- Find \(\frac{du}{dx}\):
\( \frac{du}{dx} = \cos x \)
- Identify the Outer Function (y in terms of u): Replace the inner part with \( u \).
\( y = e^u \)
- Find \(\frac{dy}{du}\):
\( \frac{dy}{du} = e^u \)
- Apply the Chain Rule: Multiply the derivatives, replacing \( u \) with \( \sin x \) at the end.
\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = (e^u) \times (\cos x) = e^{\sin x} \cos x \]
Tip for Speed: Always differentiate the outside structure first, keeping the inside exactly the same. THEN multiply by the derivative of the inside.
Example: If \( y = \ln(4x^2+1) \), Outside is \(\ln(\dots)\), Inside is \(4x^2+1\).
- Outside derivative: \( \frac{1}{\text{inside}} \). So, \( \frac{1}{4x^2+1} \).
- Inside derivative: \( 8x \).
- Result: \( \frac{dy}{dx} = \frac{1}{4x^2+1} \times 8x = \frac{8x}{4x^2+1} \).
Key Takeaway for Chain Rule: It is used whenever you have a function nested inside another function. "Differentiate the outside, multiply by the derivative of the inside."
3. The Product Rule
The Product Rule is used when you need to differentiate the product of two separate functions of x. For instance, \( y = x^2 \cos x \) or \( y = e^x \ln x \). You cannot simply differentiate each part separately!
3.1. The Product Rule Formula
If \( y = uv \), where \( u \) and \( v \) are functions of \( x \), then:
\[ \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \]
Mnemonic: Think of it as: U D-V plus V D-U.
3.2. Step-by-Step Application
Goal: Differentiate \( y = x^3 e^{5x} \)
- Identify \( u \) and \( v \):
\( u = x^3 \)
\( v = e^{5x} \)
- Find the derivatives (\(\frac{du}{dx}\) and \(\frac{dv}{dx}\)):
\( \frac{du}{dx} = 3x^2 \)
\( \frac{dv}{dx} = 5e^{5x} \) (Chain Rule applied here!)
- Substitute into the formula (\( u\frac{dv}{dx} + v\frac{du}{dx} \)):
\[ \frac{dy}{dx} = (x^3)(5e^{5x}) + (e^{5x})(3x^2) \]
- Simplify (Always factorise if possible!):
\[ \frac{dy}{dx} = 5x^3 e^{5x} + 3x^2 e^{5x} = x^2 e^{5x} (5x + 3) \]
A common mistake is thinking the derivative of \( uv \) is just \( u'v' \). It is not! You must use the full product rule formula.
Key Takeaway for Product Rule: Use it when you are multiplying two distinct variables functions together. The order of the two terms ($u\frac{dv}{dx}$ and $v\frac{du}{dx}$) doesn't matter because you are adding them.
4. The Quotient Rule
The Quotient Rule is used when you need to differentiate the division of two separate functions of x. For example, \( y = \frac{\sin x}{x^2+1} \).
4.1. The Quotient Rule Formula
If \( y = \frac{u}{v} \), where \( u \) (Numerator) and \( v \) (Denominator) are functions of \( x \), then:
\[ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]
Mnemonic (This is essential for the Quotient Rule):
"Low D-High minus High D-Low, over Low-Squared."
(Where Low = \( v \), High = \( u \), D-High = \(\frac{du}{dx}\), etc.)
The order of subtraction (Low D-High minus High D-Low) is vital!
4.2. Step-by-Step Application
Goal: Differentiate \( y = \frac{3x}{\ln x} \)
- Identify \( u \) (High) and \( v \) (Low):
\( u = 3x \)
\( v = \ln x \)
- Find the derivatives:
\( \frac{du}{dx} = 3 \)
\( \frac{dv}{dx} = \frac{1}{x} \)
- Substitute into the formula (\( \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \)):
\[ \frac{dy}{dx} = \frac{(\ln x)(3) - (3x)(\frac{1}{x})}{(\ln x)^2} \]
- Simplify: (Note that \( 3x \times \frac{1}{x} = 3 \))
\[ \frac{dy}{dx} = \frac{3\ln x - 3}{(\ln x)^2} = \frac{3(\ln x - 1)}{(\ln x)^2} \]
Did you know? You can use the Product Rule instead of the Quotient Rule if you rewrite the quotient using negative indices. For example, \( \frac{x}{e^x} = x e^{-x} \). However, the Quotient Rule is generally safer and faster for complicated denominators.
Key Takeaway for Quotient Rule: Use the "Low D-High minus High D-Low, over Low-Squared" mnemonic to remember the formula correctly, especially the subtraction order.
5. Implicit Differentiation
So far, all functions have been explicit, meaning \( y \) is clearly expressed in terms of \( x \) (e.g., \( y = x^2 + 5 \)).
Implicit functions are those where \( x \) and \( y \) are mixed up and cannot easily be separated, such as \( x^2 + y^2 = 25 \) (a circle) or \( \sin(y) + xy = e^x \).
When we differentiate an implicit function with respect to \( x \), every term containing \( y \) must be differentiated using the Chain Rule.
5.1. The Key Rule: Differentiating Y Terms
When differentiating any term involving \( y \) with respect to \( x \), you differentiate the term as normal (as if the variable was $x$) and then immediately multiply by \( \frac{dy}{dx} \).
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Example: Differentiate \( y^2 \) with respect to \( x \).
\( \frac{d}{dx} (y^2) = 2y \cdot \frac{dy}{dx} \)
-
Example: Differentiate \( \sin(y) \) with respect to \( x \).
\( \frac{d}{dx} (\sin y) = \cos y \cdot \frac{dy}{dx} \)
5.2. Step-by-Step Implicit Differentiation
Goal: Find \( \frac{dy}{dx} \) for the curve \( x^2 + 4y^3 = 5xy \)
- Differentiate Term-by-Term:
-
LHS, Term 1 (\( x^2 \)): Differentiate normally.
\( \frac{d}{dx} (x^2) = 2x \)
-
LHS, Term 2 (\( 4y^3 \)): Differentiate as normal, multiply by \( \frac{dy}{dx} \).
\( \frac{d}{dx} (4y^3) = 12y^2 \frac{dy}{dx} \)
-
RHS, Term 3 (\( 5xy \)): This is a Product Rule! Let \( u=5x \) and \( v=y \).
\( \frac{du}{dx} = 5 \)
\( \frac{dv}{dx} = 1 \cdot \frac{dy}{dx} \)
Product Rule Result: \( v\frac{du}{dx} + u\frac{dv}{dx} = y(5) + 5x(\frac{dy}{dx}) = 5y + 5x \frac{dy}{dx} \)
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LHS, Term 1 (\( x^2 \)): Differentiate normally.
- Equate the Derivatives:
\[ 2x + 12y^2 \frac{dy}{dx} = 5y + 5x \frac{dy}{dx} \]
- Collect Terms and Factorise: Move all terms containing \( \frac{dy}{dx} \) to one side and everything else to the other.
\[ 12y^2 \frac{dy}{dx} - 5x \frac{dy}{dx} = 5y - 2x \]
Factorise out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} (12y^2 - 5x) = 5y - 2x \]
- Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{5y - 2x}{12y^2 - 5x} \]
Important Note: When finding the gradient at a specific point \((x_1, y_1)\) for an implicit curve, you will need to substitute both the \(x\) and \(y\) coordinates into the final derivative expression.
Key Takeaway for Implicit Differentiation: Always remember the Chain Rule contribution: whenever you differentiate a \(y\) term, you must include the factor \( \frac{dy}{dx} \).
6. Applications: Tangents, Normals, and Rates of Change
The main purpose of finding the derivative \( \frac{dy}{dx} \) is finding the gradient (slope) of a curve at a specific point. This is crucial for defining tangents and normals.
6.1. Tangents and Normals
If the gradient of a curve at a point \( P(x_1, y_1) \) is \( m_T \) (the gradient of the tangent):
- The equation of the Tangent line is: \( y - y_1 = m_T (x - x_1) \).
- The gradient of the Normal line, \( m_N \), is perpendicular to the tangent: \( m_N = -\frac{1}{m_T} \).
- The equation of the Normal line is: \( y - y_1 = m_N (x - x_1) \).
6.2. Related Rates of Change
Often, quantities change over time, $t$. P3 problems often link the rates of change of two or more variables using the Chain Rule.
Suppose you know how the area ($A$) of a balloon changes with respect to its radius ($r$), $\frac{dA}{dr}$, and you know how the radius changes with respect to time ($t$), $\frac{dr}{dt}$. To find how the area changes with respect to time, $\frac{dA}{dt}$, we use:
\[ \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} \]
This is just the Chain Rule applied to rates! You need to set up the appropriate chain that links the rate you know with the rate you want to find.
Example 6.2: Related Rates Setup
If water is filling a cylindrical tank such that the volume $V$ changes at a rate of $0.5 \text{ m}^3/\text{min}$ (\( \frac{dV}{dt} = 0.5 \)), and we want to find the rate at which the height $h$ is changing (\( \frac{dh}{dt} \)), we need a formula linking $V$ and $h$ (\( V = \pi r^2 h \)) and then use the chain:
\[ \frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt} \]
(Note that \(\frac{dh}{dV}\) is the reciprocal of \(\frac{dV}{dh}\).)
Differentiation techniques require practice, practice, practice! If you can master identifying when to use the Chain, Product, or Quotient Rules, and handle the implicit differentiation mechanics, you have conquered the hardest part of P3!