🌟 Further Pure Mathematics 2: Comprehensive Differentiation Notes (9231) 🌟
Hello future Further Mathematician! Welcome to Differentiation in FP2. You’ve already mastered the foundational rules, but here we dive into powerful techniques needed for complex functions and series expansions. This chapter is vital for linking concepts across calculus, particularly when moving into differential equations and series analysis. Don't worry if the formulae look intimidating—they are often provided, and the key is mastering the underlying techniques!
2.3 Differentiation: Handling the Advanced World
1. Differentiation of Hyperbolic and Inverse Hyperbolic Functions
In Further Mathematics, we frequently work with hyperbolic functions (sinh, cosh, tanh) and their inverses. While related to standard trigonometric functions, their derivatives have crucial sign differences you must remember!
a) Derivatives of Standard Hyperbolic Functions
Recall the definitions:
\(\sinh x = \frac{e^x - e^{-x}}{2}\)
\(\cosh x = \frac{e^x + e^{-x}}{2}\)
The derivatives are surprisingly simple:
- Derivative of sinh x:
\(\frac{d}{dx} (\sinh x) = \cosh x\) - Derivative of cosh x:
\(\frac{d}{dx} (\cosh x) = \sinh x\) - Derivative of tanh x:
\(\frac{d}{dx} (\tanh x) = \text{sech}^2 x\)
🧠 Memory Aid: Notice the symmetry! Unlike trigonometric functions where differentiating cosine gives negative sine, differentiating cosh gives positive sinh. Hyperbolic functions are generally "positive" about differentiation.
b) Derivatives of Inverse Hyperbolic Functions
These derivatives often appear when solving integrals (as you’ll see in the Integration chapter). You should be able to differentiate them and recognise the standard forms.
- Derivative of sinh\(^{-1}\)x:
\(\frac{d}{dx} (\sinh^{-1} x) = \frac{1}{\sqrt{1+x^2}}\) - Derivative of cosh\(^{-1}\)x:
\(\frac{d}{dx} (\cosh^{-1} x) = \frac{1}{\sqrt{x^2-1}}\) - Derivative of tanh\(^{-1}\)x:
\(\frac{d}{dx} (\tanh^{-1} x) = \frac{1}{1-x^2}\)
Did you know? You can derive these derivative formulas easily using the Chain Rule and the logarithmic definitions of the inverse hyperbolic functions (e.g., \(\sinh^{-1}x = \ln(x + \sqrt{x^2+1})\)).
Key Takeaway:
Master the six core hyperbolic derivatives (including the inverses) and pay close attention to the indices and signs, especially the positive result for \(\frac{d}{dx} (\cosh x)\).
2. Obtaining the Second Derivative (\(\frac{d^2y}{dx^2}\))
In Further Maths, simply finding \(\frac{dy}{dx}\) is often just the first step. Finding the second derivative becomes crucial, especially when the relationship between \(x\) and \(y\) is complex (implicit) or defined via a third variable (parametric).
a) Second Derivative using Implicit Differentiation
This is needed when \(x\) and \(y\) are related in an equation but \(y\) is not explicitly isolated (e.g., \(x^2 + y^2 = 9\)).
Step-by-Step Process:
- Find \(\frac{dy}{dx}\): Differentiate the entire equation with respect to \(x\). Remember that any term involving \(y\) must be differentiated using the Chain Rule, multiplying by \(\frac{dy}{dx}\).
- Rearrange for \(\frac{dy}{dx}\): Isolate the expression for the first derivative.
- Find \(\frac{d^2y}{dx^2}\): Differentiate the expression for \(\frac{dy}{dx}\) again with respect to \(x\). This step almost always requires the Product Rule or the Quotient Rule.
- Substitute: Every time you differentiate a \(y\) term, a \(\frac{dy}{dx}\) factor appears. You must replace all instances of \(\frac{dy}{dx}\) in the final expression for \(\frac{d^2y}{dx^2}\) using the result from Step 2. This ensures your final answer for \(\frac{d^2y}{dx^2}\) only contains \(x\) and \(y\) (and constants).
🚨 Common Mistake: Forgetting to substitute the initial expression for \(\frac{dy}{dx}\) back into the final result for the second derivative. The answer MUST only involve \(x\) and \(y\).
Example: If \(\frac{dy}{dx} = \frac{y}{x}\), then differentiating again using the Quotient Rule gives:
\(\frac{d^2y}{dx^2} = \frac{x(\frac{dy}{dx}) - y(1)}{x^2}\)
Now substitute \(\frac{dy}{dx} = \frac{y}{x}\) back in:
\(\frac{d^2y}{dx^2} = \frac{x(\frac{y}{x}) - y}{x^2} = \frac{y - y}{x^2} = 0\)
b) Second Derivative using Parametric Differentiation
This is needed when \(x\) and \(y\) are defined in terms of a parameter, \(t\) (or \(\theta\)).
Step-by-Step Process:
- Find \(\frac{dy}{dx}\): Use the standard parametric formula:
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). This result, \(\frac{dy}{dx}\), will be a function of \(t\). Let's call this \(Z(t)\). - Find \(\frac{dZ}{dt}\): Differentiate the result \(Z(t)\) (which is \(\frac{dy}{dx}\)) with respect to the parameter \(t\).
- Find \(\frac{d^2y}{dx^2}\): Use the Chain Rule for the second derivative:
\(\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dt} \left(\frac{dy}{dx}\right) \times \frac{dt}{dx}\) - Calculate \(\frac{dt}{dx}\): Remember that \(\frac{dt}{dx} = \frac{1}{dx/dt}\).
Final Formula for Parametric Second Derivative:
\(\frac{d^2y}{dx^2} = \frac{d}{dt} \left(\frac{dy}{dx}\right) \div \frac{dx}{dt}\)
Roller Coaster Analogy: If \(\frac{dy}{dx}\) tells you the slope of the track, \(\frac{d^2y}{dx^2}\) tells you how fast that slope is changing as you move forward. When using parametric forms, you first calculate how the slope changes with respect to time (\(\frac{dZ}{dt}\)), and then you correct it by scaling it by \(\frac{dt}{dx}\) to account for speed along the x-axis.
Key Takeaway:
Implicit differentiation requires substituting \(\frac{dy}{dx}\) back into the final expression. Parametric differentiation requires dividing the derivative of the first derivative (with respect to \(t\)) by \(\frac{dx}{dt}\).
3. Maclaurin's Series and Successive Differentiation
The Maclaurin’s series provides a polynomial approximation of a function \(f(x)\) around \(x=0\).
Maclaurin’s Series Definition (First few terms):
\(f(x) \approx f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \dots\)
The syllabus requires you to derive and use the first few terms (typically up to \(x^3\) or \(x^4\)).
a) Finding Terms via Direct Successive Differentiation
For standard functions (like \(e^{\sin x}\) or \(\ln(1+x)\)), this involves direct calculation:
- Evaluate \(f(0)\).
- Differentiate once to find \(f'(x)\), then evaluate \(f'(0)\).
- Differentiate again to find \(f''(x)\), then evaluate \(f''(0)\).
- Continue as necessary (e.g., \(f'''(0)\)).
- Substitute these values into the Maclaurin formula.
💡 Tip for Complex Products/Chains: Be methodical! Write out each derivative clearly and simplify before differentiating the next time. The Chain Rule, Product Rule, and Quotient Rule are your best friends here.
b) Finding Terms using Implicit Differentiation (Advanced Maclaurin)
Sometimes, successive differentiation of \(f'(x)\) can become algebraically very complicated. If the function satisfies a differential equation, or if it has a simple derivative that relates back to the original function, we can use implicit differentiation to find the series coefficients more easily.
Example: Find the Maclaurin expansion for \(y = \tan x\).
- Find \(y\) at \(x=0\): \(y(0) = \tan(0) = 0\).
- Find \(y'\): \(y' = \sec^2 x = 1 + \tan^2 x = 1 + y^2\).
\(y'(0) = 1 + (0)^2 = 1\). - Find \(y''\) (Differentiate implicitly):
Differentiate \(y' = 1 + y^2\) with respect to \(x\):
\(y'' = 0 + 2y \left(\frac{dy}{dx}\right) = 2y y'\).
\(y''(0) = 2(0)(1) = 0\). - Find \(y'''\): Use the Product Rule on \(y'' = 2y y'\):
\(y''' = 2 \left[ y' \cdot y' + y \cdot y'' \right] = 2 \left[ (y')^2 + y y'' \right]\).
\(y'''(0) = 2 \left[ (1)^2 + (0)(0) \right] = 2\). - Substitute into Series:
\(f(x) = 0 + x(1) + \frac{x^2}{2!}(0) + \frac{x^3}{3!}(2) + \dots\)
\(\tan x \approx x + \frac{2x^3}{6} = x + \frac{x^3}{3}\)
This method allows you to find higher derivatives without getting stuck in complex trigonometric identities.
Quick Review Box: Differentiation Techniques
- Hyperbolic Functions: Differentiate just like trig functions, but \(\cosh x\) stays positive!
- Implicit Second Derivative: Differentiate twice, then substitute \(\frac{dy}{dx}\) back in.
- Parametric Second Derivative: \(\frac{d^2y}{dx^2} = \frac{d}{dt} (\frac{dy}{dx}) \times \frac{dt}{dx}\).
- Maclaurin’s Series: Calculate \(f(0), f'(0), f''(0), \dots\) either directly or using implicit successive differentiation (if the initial derivative simplifies the process).
Key Takeaway:
Maclaurin's series tests your ability to perform sequential differentiation accurately. When working with complex functions or implicit relations, remember that differentiating the first derivative implicitly is often the fastest path to the required coefficient.