# Differentiation (FP3)

> Edexcel International A-Level Further Mathematics · Edexcel IAL FMATHS FP3
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp3-differentiation/

This guide covers differentiation of hyperbolic functions, and inverse trigonometric / inverse hyperbolic functions for Edexcel IAL Further Maths FP3, with worked examples and exam tips.

**Prerequisites:** [Edexcel IAL Pure Maths differentiation rules (chain, product, quotient)](https://www.owlsprep.com/study/edexcel-ial-pure-math-p3-differentiation/); [FP3 Hyperbolic Functions definitions and notation](https://www.owlsprep.com/study/edexcel-ial-further-math-fp3-hyperbolic-functions/)

## Learning objectives

- Differentiate expressions involving hyperbolic functions using chain, product and quotient rules
- Differentiate inverse trigonometric and inverse hyperbolic functions combined with standard differentiation rules
- Simplify surd and rational differentiation results fully to meet exam marking criteria

## Differentiating Hyperbolic Functions

You will use the standard derivatives of hyperbolic functions (given in your formula book) combined with core pure maths differentiation rules to evaluate derivatives of expressions involving $\sinh x$, $\cosh x$ and $\tanh x$.

$$\frac{d}{dx}\left(\sinh x\right) = \cosh x, \quad \frac{d}{dx}\left(\cosh x\right) = \sinh x, \quad \frac{d}{dx}\left(\tanh x\right) = \text{sech}^2 x$$

> **tip**
>
> You do not need to derive these standard results, but you should recognise them instantly to speed up exam working.

**Worked example:** Differentiate $f(x) = x \sinh^2 3x$

1. Recognise the function is a product of $u=x$ and $v=\sinh^2 3x$, so apply the product rule: $f'(x) = u'v + uv'$
2. Calculate $u'$: $u' = \frac{d}{dx}(x) = 1$
3. Calculate $v'$ using the chain rule twice: let $w=3x$, so $v=(\sinh w)^2$, so $v' = 2\sinh w \cdot \cosh w \cdot 3 = 6\sinh 3x \cosh 3x$
4. Substitute into the product rule: $f'(x) = \sinh^2 3x + 6x \sinh 3x \cosh 3x$
5. Factor the final result: $f'(x) = \sinh 3x \left(\sinh 3x + 6x \cosh 3x\right)$

> **Exam tip:** Always factor out common terms from your final result to meet simplification requirements for full marks.

*Calculator:* allowed

## Differentiating Inverse Trigonometric Functions

Standard derivatives of inverse trigonometric functions are provided in your formula book. Combine these with core differentiation rules to evaluate derivatives of more complex inverse trig expressions.

$$\frac{d}{dx}\left(\arcsin x\right) = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\left(\arccos x\right) = -\frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\left(\arctan x\right) = \frac{1}{1+x^2}$$

> **warning**
>
> The negative sign in the derivative of $\arccos x$ is a common source of lost marks, so always double-check this when working.

**Worked example:** Differentiate $g(x) = \arcsin x + x\sqrt{1-x^2}$

1. Split the function into two terms to differentiate individually: $g'(x) = \frac{d}{dx}(\arcsin x) + \frac{d}{dx}\left(x\sqrt{1-x^2}\right)$
2. Differentiate the first term using the standard result: $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$
3. Apply the product rule to the second term: let $u=x$, $v=(1-x^2)^{1/2}$, so $u'=1$, $v' = -\frac{x}{\sqrt{1-x^2}}$
4. Evaluate the second term derivative: $\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}$
5. Combine and simplify all terms: $g'(x) = \frac{1 + (1-x^2) - x^2}{\sqrt{1-x^2}} = 2\sqrt{1-x^2}$

> **Exam tip:** Combine terms with square root denominators into a single fraction before simplifying to avoid arithmetic errors.

*Calculator:* allowed

## Differentiating Inverse Hyperbolic Functions

Standard derivatives of inverse hyperbolic functions are also provided in your formula book. Apply the same chain, product and quotient rules you use for other function types to these expressions.

$$\frac{d}{dx}\left(\text{arsinh }x\right) = \frac{1}{\sqrt{1+x^2}}, \quad \frac{d}{dx}\left(\text{arcosh }x\right) = \frac{1}{\sqrt{x^2-1}}, \quad \frac{d}{dx}\left(\text{artanh }x\right) = \frac{1}{1-x^2}$$

**Worked example:** Differentiate $h(x) = \text{artanh}(x^2)$

1. Recognise this is a composite function, so apply the chain rule: let $u=x^2$, so $h(x) = \text{artanh }u$
2. Apply the chain rule formula: $h'(x) = \frac{d}{du}(\text{artanh }u) \cdot \frac{du}{dx}$
3. Substitute the standard derivative: $\frac{d}{du}(\text{artanh }u) = \frac{1}{1-u^2}$
4. Calculate $\frac{du}{dx} = 2x$
5. Substitute back and simplify: $h'(x) = \frac{2x}{1-x^4}$

> **Exam tip:** Always confirm the domain of the inverse function when evaluating derivatives, e.g. $\text{arcosh }x$ is only defined for $x\geq1$.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting the negative sign in the derivative of $\arccos x$
  - Why it fails: Mixing up the signs of $\arcsin x$ and $\arccos x$ derivatives
  - Correct: Double-check the sign for $\arccos x$ and cross-reference the formula book if unsure
- **Wrong:** Differentiating $\sinh^2 x$ as $2\sinh x$ instead of $2\sinh x \cosh x$
  - Why it fails: Missing the inner derivative of the hyperbolic function when applying the chain rule
  - Correct: Apply the chain rule twice for powers of hyperbolic functions: once for the power, once for the hyperbolic term
- **Wrong:** Leaving results with square root denominators unsimplified
  - Why it fails: Exam mark schemes require fully simplified results for full marks
  - Correct: Combine fractions with surd denominators, cancel common factors and rationalise before writing your final answer
- **Wrong:** Writing $1/\sqrt{1-x^2}$ as the derivative of $\text{arsinh }x$ instead of $1/\sqrt{1+x^2}$
  - Why it fails: Mixing up inverse trig and inverse hyperbolic derivative forms
  - Correct: Remember inverse trig derivatives have $1-x^2$ under the root, while $\text{arsinh }x$ has $1+x^2$
- **Wrong:** Differentiating $\tanh 3x$ as $\text{sech}^2 3x$ instead of $3\text{sech}^2 3x$
  - Why it fails: Rushing through chain rule steps for linear inner functions
  - Correct: Write the inner function explicitly when applying the chain rule to avoid missing its derivative

## Cheatsheet

| Function | Standard Derivative | Key Tip |
| --- | --- | --- |
| $\sinh x$ | $\cosh x$ | No sign change |
| $\cosh x$ | $\sinh x$ | No sign change |
| $\tanh x$ | $\text{sech}^2 x$ | Squared hyperbolic secant |
| $\arcsin x$ | $1/\sqrt{1-x^2}$ | Positive denominator |
| $\arccos x$ | $-1/\sqrt{1-x^2}$ | Negative numerator |
| $\arctan x$ | $1/(1+x^2)$ | No square root |
| $\text{arsinh }x$ | $1/\sqrt{1+x^2}$ | Plus sign under root |
| $\text{arcosh }x$ | $1/\sqrt{x^2-1}$ | $x^2-1$ under root |
| $\text{artanh }x$ | $1/(1-x^2)$ | No square root, $\|x\|<1$ |

## What's next

Now you have mastered FP3 differentiation of hyperbolic and inverse functions, you are ready to move on to FP3 Integration, which uses the reverse of the derivative rules you have learned here. This differentiation knowledge will also form the foundation for solving differential equations in FP3, as well as calculus applications in other Further Maths units like FP2 and Mechanics 3. Make sure you practice plenty of past paper questions on this topic to build speed and accuracy, as differentiation questions are often combined with other topics in longer exam questions worth 5-8 marks. Regular practice will help you avoid the common pitfalls listed earlier and ensure you can simplify results quickly under exam conditions.

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