Edexcel IAL - Further Pure Mathematics 3- 3.2 Differentiation of Inverse Trigonometric and Hyperbolic Functions- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 3- 3.2 Differentiation of Inverse Trigonometric and Hyperbolic Functions -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 3- 3.2 Differentiation of Inverse Trigonometric and Hyperbolic Functions -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 3.2 Differentiation of Inverse Trigonometric and Hyperbolic Functions
Differentiation of Inverse Functions
Inverse functions arise frequently in calculus, particularly inverse trigonometric and inverse hyperbolic functions. Differentiation usually involves the chain rule and careful algebraic simplification.
Key Derivatives of Inverse Trigonometric Functions
| Function | Derivative |
|---|---|
| \( \arcsin x \) | \( \dfrac{1}{\sqrt{1-x^2}} \) |
| \( \arccos x \) | \( -\dfrac{1}{\sqrt{1-x^2}} \) |
| \( \arctan x \) | \( \dfrac{1}{1+x^2} \) |
Key Derivatives of Inverse Hyperbolic Functions
| Function | Derivative |
|---|---|
| \( \operatorname{arsinh} x \) | \( \dfrac{1}{\sqrt{1+x^2}} \) |
| \( \operatorname{arcosh} x \) | \( \dfrac{1}{\sqrt{x^2-1}} \) |
| \( \operatorname{artanh} x \) | \( \dfrac{1}{1-x^2} \) |
Important Technique
When differentiating expressions involving inverse functions:
- Apply the chain rule if the argument is not simply \( x \)
- Use the product rule where necessary
- Simplify square roots carefully
Example :
Differentiate
\( y = \arcsin x + x\sqrt{1-x^2} \).
▶️ Answer/Explanation
Differentiate each term separately.
First term:
\( \dfrac{d}{dx}(\arcsin x) = \dfrac{1}{\sqrt{1-x^2}} \)
Second term uses the product rule:
\( \dfrac{d}{dx}\!\left(x\sqrt{1-x^2}\right) = \sqrt{1-x^2} + x\!\left(-\dfrac{x}{\sqrt{1-x^2}}\right) \)
\( = \dfrac{1-2x^2}{\sqrt{1-x^2}} \)
Adding results:
\( \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1-x^2}} + \dfrac{1-2x^2}{\sqrt{1-x^2}} \)
\( = 2\sqrt{1-x^2} \)
Conclusion: \( \dfrac{dy}{dx} = 2\sqrt{1-x^2} \).
Example :
Differentiate
\( y = \dfrac{1}{2}\operatorname{artanh}(x^2) \).
▶️ Answer/Explanation
Using the chain rule:
\( \dfrac{d}{dx}[\operatorname{artanh}(x^2)] = \dfrac{1}{1-x^4}\cdot 2x \)
Including the constant factor:
\( \dfrac{dy}{dx} = \dfrac{1}{2}\cdot \dfrac{2x}{1-x^4} = \dfrac{x}{1-x^4} \)
Conclusion: \( \dfrac{dy}{dx} = \dfrac{x}{1-x^4} \).
Example :
Differentiate
\( y = x\,\operatorname{arsinh} x \).
▶️ Answer/Explanation
Use the product rule.
Let
\( u = x,\quad v = \operatorname{arsinh} x \)
Then
\( \dfrac{du}{dx} = 1 \)
\( \dfrac{dv}{dx} = \dfrac{1}{\sqrt{1+x^2}} \)
Applying the product rule:
\( \dfrac{dy}{dx} = \operatorname{arsinh} x + \dfrac{x}{\sqrt{1+x^2}} \)
Conclusion: The derivative is \( \operatorname{arsinh} x + \dfrac{x}{\sqrt{1+x^2}} \).