Further Mathematics | Differentiation

Welcome to FP3 Differentiation! Mastering New Functions

Hello future Mathematician! Welcome to the Differentiation chapter of Further Pure Mathematics 3 (FP3). Don't panic! We are simply building on the powerful techniques you mastered in P3 (Chain Rule, Product Rule, Implicit Differentiation).

In this chapter, we significantly expand your toolkit by learning how to differentiate two new families of functions:
1. Inverse Trigonometric Functions (like \(\arcsin x\)).
2. Hyperbolic Functions and their Inverses (like \(\cosh x\) and \(\operatorname{arsinh} x\)).

Mastering these techniques is crucial, as the results form the foundation for many complex integrations and later chapters involving series expansions and differential equations. Let's dive in!

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Section 1: Your Differentiation Toolkit Review (The Essentials)

Before we tackle the new functions, let's briefly review the most important rule you'll need constantly: The Chain Rule.

Review: The Chain Rule

The Chain Rule is used when differentiating a function *of* a function. If \(y = f(u)\) and \(u = g(x)\), then:

$$ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} $$

Example: If \(y = \cosh(3x^2)\), we let \(u = 3x^2\). Then \(\frac{dy}{du} = \sinh u\) and \(\frac{du}{dx} = 6x\).
Therefore, \(\frac{dy}{dx} = (\sinh(3x^2)) \times (6x) = 6x \sinh(3x^2)\). (We’ll cover \(\cosh\) differentiation properly soon, but the method is the same!)

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Section 2: Differentiating Inverse Trigonometric Functions

The inverse trigonometric functions (\(\arcsin x\), \(\arccos x\), \(\arctan x\)) answer the question: "What angle gives me this ratio?"

In FP3, we don't just use these functions; we derive their derivatives using Implicit Differentiation and the standard identities you learned in P3.

2.1: Deriving the Derivative of \(y = \arcsin x\)

Don't worry about memorizing the derivation, but understanding the steps helps solidify the result!

1. Start with \(y = \arcsin x\). This means \(x = \sin y\).
2. Differentiate implicitly with respect to \(x\): $$ \frac{d}{dx}(x) = \frac{d}{dx}(\sin y) $$ $$ 1 = (\cos y) \frac{dy}{dx} $$
3. Rearrange to find \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = \frac{1}{\cos y} $$
4. Use the P3 identity \(\sin^2 y + \cos^2 y = 1\), which means \(\cos y = \sqrt{1 - \sin^2 y}\).
5. Since \(x = \sin y\), substitute \(x\) back in: $$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} $$

2.2: The Standard Results (Memorize These!)

These are essential formulas you will apply directly. Notice the beautiful similarity!

Inverse Trigonometric Derivatives

If \(|x| < 1\):

• $$ \frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}} $$
• $$ \frac{d}{dx}(\arccos x) = - \frac{1}{\sqrt{1 - x^2}} $$

For all real \(x\):

• $$ \frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2} $$

2.3: Applying the Chain Rule to Inverse Trig Functions

If you have a function of \(x\) inside the inverse trig function, apply the Chain Rule.

Example: Find \(\frac{dy}{dx}\) for \(y = \arcsin(4x^3)\).

1. Identify the outer function: \(\arcsin u\). The inner function is \(u = 4x^3\).
2. Differentiate the inner function: \(\frac{du}{dx} = 12x^2\).
3. Differentiate the outer function with respect to \(u\) (using the standard formula): \(\frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}\).
4. Combine: $$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (4x^3)^2}} \times (12x^2) = \frac{12x^2}{\sqrt{1 - 16x^6}} $$

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Section 3: Differentiating Hyperbolic Functions

Hyperbolic functions (\(\sinh x\), \(\cosh x\), \(\tanh x\)) are based on the exponent \(e^x\). They behave similarly to circular trig functions but have crucial differences in their derivatives.

3.1: Hyperbolic Standard Derivatives

Remember the definitions: \(\sinh x = \frac{e^x - e^{-x}}{2}\) and \(\cosh x = \frac{e^x + e^{-x}}{2}\). You can derive their derivatives easily:

$$ \frac{d}{dx}(\sinh x) = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} = \cosh x $$

Hyperbolic Derivatives (The Positive Ones!)

• $$ \frac{d}{dx}(\sinh x) = \cosh x $$
• $$ \frac{d}{dx}(\cosh x) = \sinh x $$
• $$ \frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x $$

Key Takeaway: Notice that unlike circular trig, differentiating \(\cosh x\) does NOT introduce a minus sign! This is a critical point where students often make mistakes.

3.2: Differentiating Hyperbolic Reciprocals

You should also know the derivatives of the other hyperbolic functions:

• $$ \frac{d}{dx}(\operatorname{cosech} x) = -\operatorname{cosech} x \coth x $$
• $$ \frac{d}{dx}(\operatorname{sech} x) = -\operatorname{sech} x \tanh x $$
• $$ \frac{d}{dx}(\coth x) = -\operatorname{cosech}^2 x $$

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Section 4: Differentiating Inverse Hyperbolic Functions

This section contains some of the most complex standard results in FP3. The inverse hyperbolic functions (\(\operatorname{arsinh} x, \operatorname{arcosh} x, \operatorname{artanh} x\)) can be defined using logarithms (which you derived in the Identities chapter). We use these log forms to find the derivatives.

4.1: Derivation Example: \(y = \operatorname{arsinh} x\)

We could derive this using implicit differentiation (like we did for \(\arcsin x\)), or we can use the log definition:
$$ y = \operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1}) $$

Differentiating this using the Chain Rule (and being very careful with the bracket inside the \(\ln\)):

$$ \frac{dy}{dx} = \frac{1}{\text{inside}} \times \frac{d}{dx}(\text{inside}) $$

After simplification (which requires differentiating the square root term, \(\frac{d}{dx}(\sqrt{x^2 + 1})\)), the final result is extremely clean:

4.2: The Standard Inverse Hyperbolic Results (Crucial!)

These results must be committed to memory, as they are used constantly in integration as well.

Inverse Hyperbolic Derivatives

For all real \(x\):

• $$ \frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2 + 1}} $$

For \(x > 1\):

• $$ \frac{d}{dx}(\operatorname{arcosh} x) = \frac{1}{\sqrt{x^2 - 1}} $$

For \(|x| < 1\):

• $$ \frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1 - x^2} $$

4.3: Example Application (Chain Rule with Inverse Hyperbolic)

Find \(\frac{dy}{dx}\) for \(y = \operatorname{arcosh}(\sec x)\).

1. Identify \(u = \sec x\). \(\frac{du}{dx} = \sec x \tan x\).
2. Apply the \(\operatorname{arcosh}\) formula to \(u\): $$ \frac{dy}{du} = \frac{1}{\sqrt{u^2 - 1}} = \frac{1}{\sqrt{\sec^2 x - 1}} $$
3. Use the identity \(\tan^2 x + 1 = \sec^2 x\), so \(\sec^2 x - 1 = \tan^2 x\). $$ \frac{dy}{du} = \frac{1}{\sqrt{\tan^2 x}} = \frac{1}{\tan x} $$
4. Combine using the Chain Rule: $$ \frac{dy}{dx} = \frac{1}{\tan x} \times (\sec x \tan x) $$
5. Simplify: $$ \frac{dy}{dx} = \sec x $$

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Key Takeaways and Final Encouragement

You've significantly upgraded your differentiation skills! The true difficulty in this chapter lies not in the technique (which is usually the Chain Rule), but in accurate recall of the standard results.

Summary of Differentiation Standards (FP3)

Function \(y\)	Derivative \(\frac{dy}{dx}\)	Notes
\(\arcsin x\)	\(\frac{1}{\sqrt{1 - x^2}}\)	Negative if it's \(\arccos x\)
\(\arctan x\)	\(\frac{1}{1 + x^2}\)	No square root here!
\(\sinh x\)	\(\cosh x\)	Positive!
\(\cosh x\)	\(\sinh x\)	Still Positive!
\(\operatorname{arsinh} x\)	\(\frac{1}{\sqrt{x^2 + 1}}\)	\(x^2 + 1\) (Always positive, easy domain)
\(\operatorname{arcosh} x\)	\(\frac{1}{\sqrt{x^2 - 1}}\)	\(x^2 - 1\) (Requires \(x > 1\))
\(\operatorname{artanh} x\)	\(\frac{1}{1 - x^2}\)	No square root here!

Practice makes perfect. Do plenty of examples, particularly those that require combining the Chain Rule with these new formulas. You've got this!