Differentiation

Q: What is the formula for the Chain Rule when differentiating a function of a function?

The formula is $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$, where $y$ is a function of $u$ and $u$ is a function of $x$.

Q: What are the Product and Quotient Rule formulas used for combined functions?

Product Rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$. Quotient Rule: $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Q: How is implicit differentiation used to prove the derivative of $y = \text{arcsin } x$?

Rewrite the function as $x = \text{sin } y$, differentiate implicitly to get $1 = \text{cos } y \frac{dy}{dx}$, and substitute $\text{cos } y = \text{\sqrt{1 - sin^2 y}}$ to find $\frac{dy}{dx} = \frac{1}{\text{\sqrt{1-x^2}}}$.

Q: What is the derivative of $\text{arcsin } x$ and how does it compare to $\text{arccos } x$?

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

Q: What is the standard derivative of $\text{arctan } x$?

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

Q: What are the derivatives of the primary hyperbolic functions $\text{sinh } x$ and $\text{cosh } x$?

$\frac{d}{dx}(\text{sinh } x) = \text{cosh } x$ and $\frac{d}{dx}(\text{cosh } x) = \text{sinh } x$.

Q: What is the derivative of $\text{tanh } x$?

$\frac{d}{dx}(\text{tanh } x) = \text{sech}^2 x$.

Q: What is the derivative of $\text{arsinh } x$?

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

Q: What is the derivative of $\text{arcosh } x$ and what is its domain?

$\frac{d}{dx}(\text{arcosh } x) = \frac{1}{\text{\sqrt{x^2-1}}}$, defined for $x > 1$.

Pure Mathematics (YPM01) · Pearson Edexcel International A Level · 6 min read · Updated: 6 Sept 2026

Welcome to Unit FP3: The Power of Advanced Differentiation

Hello future mathematician! This chapter, Differentiation in Further Pure 3 (FP3), is where we take the powerful tools you learned in Pure 3 (P3) and apply them to a whole new world of complex functions: the Inverse Trigonometric and the Hyperbolic functions.

Don't worry if these functions look strange! The core rules (Chain, Product, Quotient) haven't changed. We are simply expanding our toolbox to find the derivatives (the instantaneous rate of change or gradient) of these specialized equations. Mastery here is essential for success in later topics like integration and solving differential equations!

Section 1: The Core Tools (A Quick Refresher)

Before diving into new functions, let's briefly remember the key differentiation techniques you must be fluent in. These rules will always be used in combination with the new functions we introduce.

1.1 The Chain Rule: Differentiating a Function of a Function

If $y$ is a function of $u$, and $u$ is a function of $x$, then:

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

Memory Aid: "Differentiate the outside, leave the inside, then multiply by the derivative of the inside." This is crucial when dealing with functions like $\arcsin(2x)$ or $\cosh^3(x)$.

1.2 Product and Quotient Rules

Remember these structures! They are often combined with the new derivatives we learn in FP3.

Product Rule: For $y = uv$, $$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$
Quotient Rule: For $y = \frac{u}{v}$, $$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$

Key Takeaway: Differentiation in FP3 rarely involves just one rule. Expect to apply the Chain Rule frequently inside the Product or Quotient rules!

Section 2: Differentiation of Inverse Trigonometric Functions

In FP3, you must know the standard derivatives for $\arcsin x$, $\arccos x$, and $\arctan x$. We usually derive these using Implicit Differentiation, which proves why their derivatives look so unique.

2.1 Understanding the Process (The Proof for $\arcsin x$)

If $y = \arcsin x$, we rewrite it as $x = \sin y$. Now, differentiate implicitly with respect to $x$:

$\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$
$1 = \cos y \cdot \frac{dy}{dx}$
Therefore, $\frac{dy}{dx} = \frac{1}{\cos y}$

Since $\cos^2 y + \sin^2 y = 1$, we know $\cos y = \sqrt{1 - \sin^2 y}$. Substituting back $x = \sin y$:

$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$

2.2 The Essential Standard Derivatives (Memorize These!)

You must be able to state these results directly, and apply the Chain Rule when the argument is not simply $x$.

Standard Results for Inverse Trig Functions:

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

Tip for Struggling Students: Spotting the Negative

Notice that only the function starting with 'c' ($\arccos x$) has a negative derivative. This pattern is similar to standard trig differentiation ($\cos \rightarrow -\sin$).

2.3 Example Application (Using the Chain Rule)

Example: Find the derivative of $y = \arctan(3x)$.

We use the Chain Rule. Let $u = 3x$, so $\frac{du}{dx} = 3$.

Then $y = \arctan u$, so $\frac{dy}{du} = \frac{1}{1+u^2}$.

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \left(\frac{1}{1+(3x)^2}\right) \times 3 = \frac{3}{1+9x^2}$$

Common Mistake to Avoid: Forgetting to square the entire argument. It's $(3x)^2$, not $3x^2$.

Key Takeaway: Inverse trigonometric derivatives always result in a fraction involving a square root (for $\arcsin / \arccos$) or a simple squared term (for $\arctan$).

Section 3: Differentiation of Hyperbolic Functions

Hyperbolic functions ($\sinh x$, $\cosh x$, $\tanh x$) are defined using exponential functions. This means their derivatives behave slightly differently from standard trig functions—often more simply!

Did You Know? Hyperbolic functions describe the shape of a hanging chain or cable (a catenary). They are vital in engineering and physics!

3.1 The Standard Hyperbolic Derivatives

Since $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$, we can find their derivatives easily:

$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(e^{-x}) = -e^{-x}$

Standard Results for Hyperbolic Functions:

1. $\frac{d}{dx}(\sinh x) = \cosh x$
2. $\frac{d}{dx}(\cosh x) = \sinh x$
3. $\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x$
4. $\frac{d}{dx}(\operatorname{coth} x) = -\operatorname{cosech}^2 x$
5. $\frac{d}{dx}(\operatorname{sech} x) = -\operatorname{sech} x \tanh x$
6. $\frac{d}{dx}(\operatorname{cosech} x) = -\operatorname{cosech} x \coth x$

Memory Trick: The Missing Negative

When differentiating standard trig functions, $\cos x \rightarrow -\sin x$. In the hyperbolic world, this sign change is missing for the primary functions!

The derivative of $\cosh x$ is just $\sinh x$. There is NO negative sign. This is the biggest difference to memorize!

3.2 Combining Rules (Product Rule Example)

Example: Differentiate $y = x^2 \cosh(5x)$.

This requires the Product Rule $y = uv$, where $u = x^2$ and $v = \cosh(5x)$.

$\frac{du}{dx} = 2x$
$\frac{dv}{dx} = \sinh(5x) \cdot 5 = 5\sinh(5x)$ (Chain Rule used here!)

$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$

$$\frac{dy}{dx} = x^2 (5\sinh(5x)) + \cosh(5x) (2x)$$

$$\frac{dy}{dx} = 5x^2 \sinh(5x) + 2x \cosh(5x)$$

Key Takeaway: Hyperbolic differentiation is often easier than standard trig differentiation because of the absence of the negative sign when differentiating $\cosh x$. Always apply the Chain Rule!

Section 4: Differentiation of Inverse Hyperbolic Functions

These functions ($\operatorname{arsinh} x$, $\operatorname{arcosh} x$, $\operatorname{artanh} x$) are the inverse of the hyperbolic functions. They are sometimes called $\sinh^{-1} x$, $\cosh^{-1} x$, etc., but using 'ar' (area) helps distinguish them from simple reciprocals. You should be familiar with their Logarithmic Definitions, as these definitions are often used to prove the derivatives.

4.1 Proving the Derivative (Using Logarithmic Forms)

We know that $\operatorname{arsinh} x = \ln(x + \sqrt{x^2+1})$. We can differentiate this expression using the Chain Rule combined with the derivative of $\ln u$, which is $\frac{1}{u} \frac{du}{dx}$.

The result, after a lot of careful algebra, is:

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

(You may be asked to prove these results in exams, so practice the algebraic manipulation!)

4.2 The Essential Standard Derivatives (Inverse Hyperbolic)

These results look confusingly similar to the inverse trig results. Pay very close attention to the signs and whether the $x^2$ term is positive or negative within the denominator!

Standard Results for Inverse Hyperbolic Functions:

1. $\frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2+1}}$
2. $\frac{d}{dx}(\operatorname{arcosh} x) = \frac{1}{\sqrt{x^2-1}}$ (Defined for $x>1$)
3. $\frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1-x^2}$ (Defined for $|x|<1$)

A Crucial Comparison ($\arctan$ vs $\operatorname{artanh}$)

This is where many students lose marks! Notice the subtle but vital difference:

$\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$ (Note: $1$ is first)
$\frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1-x^2}$ (Note: $1$ is first, but it’s a MINUS sign)

If you confuse these, your subsequent integration work will also be wrong! Double-check the sign in the denominator.

4.3 Step-by-Step Example (Combined Chain and Inverse Hyperbolic)

Example: Find the derivative of $y = \operatorname{arcosh}(x^3 + 2)$.

Identify the form: This is $y = \operatorname{arcosh} u$, where $u = x^3 + 2$.
Differentiate the 'inside' ($u$): $\frac{du}{dx} = 3x^2$.
Apply the standard formula: $\frac{dy}{du} = \frac{1}{\sqrt{u^2-1}}$.
Combine using Chain Rule: $$\frac{dy}{dx} = \frac{1}{\sqrt{(x^3 + 2)^2 - 1}} \cdot 3x^2$$

Result: $$\frac{dy}{dx} = \frac{3x^2}{\sqrt{(x^3 + 2)^2 - 1}}$$

Accessibility Checkpoint: Don't worry if the resulting derivatives look messy. They are supposed to! The key is accurate application of the formula and the Chain Rule.

Key Takeaway: Inverse hyperbolic derivatives often involve a square root in the denominator, and the relationship between $1$ and $x^2$ dictates the sign and the function type.

Quick Review Box: The Six Core FP3 Differentiation Formulas

Keep these six formulas clearly separated in your mind!

Inverse Trig:
$\arcsin x \rightarrow \frac{1}{\sqrt{1-x^2}}$
$\arctan x \rightarrow \frac{1}{1+x^2}$

Inverse Hyperbolic:
$\operatorname{arsinh} x \rightarrow \frac{1}{\sqrt{x^2+1}}$
$\operatorname{arcosh} x \rightarrow \frac{1}{\sqrt{x^2-1}}$
$\operatorname{artanh} x \rightarrow \frac{1}{1-x^2}$

Hyperbolic:
$\cosh x \rightarrow \sinh x$ (No negative!)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the formula for the Chain Rule when differentiating a function of a function?

The formula is $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$, where $y$ is a function of $u$ and $u$ is a function of $x$.

What are the Product and Quotient Rule formulas used for combined functions?

Product Rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$.
Quotient Rule: $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

How is implicit differentiation used to prove the derivative of $y = \text{arcsin } x$?

Rewrite the function as $x = \text{sin } y$, differentiate implicitly to get $1 = \text{cos } y \frac{dy}{dx}$, and substitute $\text{cos } y = \text{\sqrt{1 - sin^2 y}}$ to find $\frac{dy}{dx} = \frac{1}{\text{\sqrt{1-x^2}}}$.

What is the derivative of $\text{arcsin } x$ and how does it compare to $\text{arccos } x$?

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

What is the standard derivative of $\text{arctan } x$?

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

What are the derivatives of the primary hyperbolic functions $\text{sinh } x$ and $\text{cosh } x$?

$\frac{d}{dx}(\text{sinh } x) = \text{cosh } x$ and $\frac{d}{dx}(\text{cosh } x) = \text{sinh } x$.

What is the derivative of $\text{tanh } x$?

$\frac{d}{dx}(\text{tanh } x) = \text{sech}^2 x$.

What is the derivative of $\text{arsinh } x$?

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

What is the derivative of $\text{arcosh } x$ and what is its domain?

$\frac{d}{dx}(\text{arcosh } x) = \frac{1}{\text{\sqrt{x^2-1}}}$, defined for $x > 1$.

How does the derivative of $\text{artanh } x$ differ from $\text{arctan } x$?

$\frac{d}{dx}(\text{artanh } x) = \frac{1}{1-x^2}$ (minus sign), while $\frac{d}{dx}(\text{arctan } x) = \frac{1}{1+x^2}$ (plus sign).

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