Edexcel IAL - Further Pure Mathematics 3- 3.1 Differentiation of Hyperbolic Functions- Study notes - New syllabus
Edexcel IAL – Further Pure Mathematics 3- 3.1 Differentiation of Hyperbolic Functions -Study notes- New syllabus
Edexcel IAL – Further Pure Mathematics 3- 3.1 Differentiation of Hyperbolic Functions -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 3.1 Differentiation of Hyperbolic Functions
Differentiation of Hyperbolic Functions
Hyperbolic functions can be differentiated using standard results. These derivatives are frequently combined with the chain rule, product rule, and quotient rule when differentiating more complicated expressions.
Basic Derivatives of Hyperbolic Functions
| Function | Derivative |
|---|---|
| \( \sinh x \) | \( \cosh x \) |
| \( \cosh x \) | \( \sinh x \) |
| \( \tanh x \) | \( \mathrm{ sech^2 x }\) |
| \( \mathrm{sech x} \) | \( -\mathrm{ sech x\,\tanh x }\) |
| \( \mathrm{csch x} \) | \( -\mathrm{ csch x\,\coth x }\) |
| \( \coth x \) | \( -\mathrm{ csch^2 x }\) |
Useful Differentiation Rules
Chain rule: If \( y = f(g(x)) \), then \( \dfrac{dy}{dx} = f'(g(x))g'(x) \)
Product rule: \( \dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \)
Quotient rule: \( \dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\dfrac{du}{dx} – u\dfrac{dv}{dx}}{v^2} \)
Differentiation of Expressions Involving Hyperbolic Functions
When differentiating expressions involving hyperbolic functions:
- Apply the chain rule when the argument is not simply \( x \)
- Use the product rule when functions are multiplied
- Use the quotient rule when functions are divided
Example :
Differentiate \( y = \tanh 3x \).
▶️ Answer/Explanation
Using the chain rule:
\( \dfrac{d}{dx}(\tanh 3x) = \sech^2(3x)\cdot 3 \)
Conclusion: \( \dfrac{dy}{dx} = 3\sech^2(3x) \).
Example :
Differentiate \( y = x\sinh^2 x \).
▶️ Answer/Explanation
Use the product rule with
\( u = x,\; v = \sinh^2 x \)
Then
\( \dfrac{du}{dx} = 1 \)
\( \dfrac{dv}{dx} = 2\sinh x\cosh x \)
Applying the product rule:
\( \dfrac{dy}{dx} = x(2\sinh x\cosh x) + \sinh^2 x \)
Conclusion: \( \dfrac{dy}{dx} = 2x\sinh x\cosh x + \sinh^2 x \).
Example :
Differentiate
\( y = \dfrac{\cosh 2x}{\sqrt{x + 1}} \).
▶️ Answer/Explanation
Use the quotient rule with
\( u = \cosh 2x,\; v = (x+1)^{1/2} \)
Then
\( \dfrac{du}{dx} = 2\sinh 2x \)
\( \dfrac{dv}{dx} = \dfrac{1}{2}(x+1)^{-1/2} \)
Applying the quotient rule:
\( \dfrac{dy}{dx} = \dfrac{(x+1)^{1/2}(2\sinh 2x) – \cosh 2x\left(\dfrac{1}{2}(x+1)^{-1/2}\right)}{x+1} \)
Conclusion: The derivative follows directly using the quotient rule.