Welcome to FP3 Differentiation!
Hi there! Welcome to one of the most exciting parts of Further Pure Mathematics 3 (FP3). You’ve already mastered differentiation in your standard Pure Mathematics units, but now we are going to level up. In this chapter, we will learn how to find the gradient of curves involving hyperbolic functions and inverse functions (both trigonometric and hyperbolic).
Don’t worry if these terms sound a bit "extra"—at its heart, this is just applying the rules you already know (like the Chain Rule and Product Rule) to some new, powerful mathematical tools. Let’s dive in!
1. Differentiating Hyperbolic Functions
Hyperbolic functions like \(\sinh x\) and \(\cosh x\) behave very similarly to standard trig functions, but with some friendly twists. The most important thing to remember is that the signs are often simpler than what you’re used to!
The "Big Three" Derivatives
You need to know these by heart:
1. \(\frac{d}{dx}(\sinh x) = \cosh x\)
2. \(\frac{d}{dx}(\cosh x) = \sinh x\)
3. \(\frac{d}{dx}(\tanh x) = \text{sech}^2 x\)
Quick Review: Remember in circular trig, \(\frac{d}{dx}(\cos x) = -\sin x\)? Well, for hyperbolic functions, there is no minus sign when differentiating \(\cosh x\). This is a very common place to lose marks, so keep an eye on it!
The Reciprocal Hyperbolic Derivatives
4. \(\frac{d}{dx}(\text{sech } x) = -\text{sech } x \tanh x\)
5. \(\frac{d}{dx}(\text{cosech } x) = -\text{cosech } x \coth x\)
6. \(\frac{d}{dx}(\text{coth } x) = -\text{cosech}^2 x\)
Memory Aid: Notice that for the reciprocal functions (\(\text{sech}, \text{cosech}, \text{coth}\)), the derivative always starts with a minus sign.
Using the Chain Rule
When you have a function inside your hyperbolic function, like \(\tanh(3x)\), you must multiply by the derivative of that inner function.
Example: Find \(\frac{dy}{dx}\) if \(y = \tanh(3x)\).
Step 1: Differentiate the "outside" (\(\tanh\)) to get \(\text{sech}^2(3x)\).
Step 2: Multiply by the derivative of the "inside" (\(3x\)), which is \(3\).
Result: \(\frac{dy}{dx} = 3\text{sech}^2(3x)\).
Key Takeaway: Hyperbolic differentiation is just like regular trig differentiation, but \(\cosh x\) differentiates to positive \(\sinh x\).
2. Differentiating Inverse Trigonometric Functions
Sometimes we need to find the rate of change for functions like \(\arcsin x\) (also written as \(\sin^{-1} x\)). Because these functions are the "undoing" of trig functions, their derivatives look quite different—they actually turn into algebraic fractions!
The Standard Results
1. \(\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}\)
2. \(\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}\)
3. \(\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}\)
Common Mistake to Avoid: Students often confuse the \(\arctan x\) derivative with \(\arcsin x\). Notice that \(\arctan x\) does not have a square root and has a plus sign!
Did you know? These formulas are incredibly useful in engineering when you need to calculate angles based on changing side lengths in a triangle.
Key Takeaway: Inverse trig derivatives result in fractions involving \(x^2\). Use the formula booklet to double-check your signs!
3. Differentiating Inverse Hyperbolic Functions
Just like their trigonometric cousins, inverse hyperbolic functions (\(\text{arsinh } x\), \(\text{arcosh } x\), and \(\text{artanh } x\)) have derivatives that are algebraic fractions. These are very common in FP3 exams.
The Formulas
1. \(\frac{d}{dx}(\text{arsinh } x) = \frac{1}{\sqrt{x^2+1}}\)
2. \(\frac{d}{dx}(\text{arcosh } x) = \frac{1}{\sqrt{x^2-1}}\) (where \(x > 1\))
3. \(\frac{d}{dx}(\text{artanh } x) = \frac{1}{1-x^2}\) (where \(|x| < 1\))
Don't worry if this seems tricky at first! You can actually derive these using a technique called Implicit Differentiation. For example, if \(y = \text{arsinh } x\), then \(x = \sinh y\). Differentiating both sides with respect to \(x\) gives \(1 = \cosh y \frac{dy}{dx}\), which leads you back to the formula!
Analogy: The "One and \(x^2\)" Puzzle
Think of these derivatives as different ways to arrange \(1\) and \(x^2\) under a square root:
- Sum (\(x^2 + 1\)) \(\rightarrow\) \(\text{arsinh}\)
- Difference, \(x\) first (\(x^2 - 1\)) \(\rightarrow\) \(\text{arcosh}\)
- Difference, \(1\) first (\(1 - x^2\)) \(\rightarrow\) \(\text{artanh}\) (no square root!)
Key Takeaway: Inverse hyperbolic derivatives are similar to inverse trig derivatives, but the signs inside the square root change. Always check if your \(x^2\) is positive or negative.
4. Combining the Rules (Complex Expressions)
In your exam, you will rarely get a simple function. You will likely see a mix of Product Rule, Quotient Rule, and Chain Rule involving these new functions.
Step-by-Step Example: \(y = x \sinh^2 x\)
This requires the Product Rule: \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\).
1. Let \(u = x\) and \(v = \sinh^2 x\).
2. Find \(\frac{du}{dx} = 1\).
3. Find \(\frac{dv}{dx}\) using the Chain Rule: \(\frac{d}{dx}((\sinh x)^2) = 2\sinh x \cosh x\).
4. Plug into the Product Rule: \(\frac{dy}{dx} = x(2\sinh x \cosh x) + (\sinh^2 x)(1)\).
5. Simplify using the identity \(2\sinh x \cosh x = \sinh 2x\).
Final Result: \(\frac{dy}{dx} = x \sinh 2x + \sinh^2 x\).
Common Pitfalls
- Power Rule Confusion: \(\sinh^2 x\) is \((\sinh x)^2\). Use the chain rule!
- Mixing Trig and Hyperbolic: \(\frac{d}{dx}(\sin x) = \cos x\), but \(\frac{d}{dx}(\sinh x) = \cosh x\). They look the same, but \(\frac{d}{dx}(\cos x) = -\sin x\) while \(\frac{d}{dx}(\cosh x) = +\sinh x\).
- Domain Restrictions: Remember that \(\text{arcosh } x\) is only defined for \(x \ge 1\). If a question asks why a derivative doesn't exist at \(x=0.5\), that's your answer!
Quick Review Box
Key Term: Hyperbolic Functions - Functions based on the geometry of a hyperbola rather than a circle.
The Most Important Rule: \(\frac{d}{dx}(\cosh x) = \sinh x\) (Positive!)
Inverse Trig vs Hyperbolic:
- \(\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}\)
- \(\frac{d}{dx}(\text{arsinh } x) = \frac{1}{\sqrt{x^2+1}}\)
Congratulations! You've covered the core differentiation concepts for FP3. Practice using these formulas with the Product and Quotient rules, and you'll be an expert in no time!