Differentiating & Integrating Hyperbolic Functions (Edexcel A Level Further Maths: Core Pure)

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Last updated

5 February 2024

$\frac{d}{d x} (\sinh x) = \cosh x$
$\frac{d}{d x} (\cosh x) = \sinh x$
$\frac{d}{d x} (\tanh x) = {sech}^{2} x$
These are given in the formulae booklet
- You can prove them by differentiating the definitions involving e
Notice that they are similar to the derivatives of the circular trig functions
- Be careful of the difference between the derivatives of cosx and coshx
  - One involves a negative sign and the other does not

The following differentiation skills may be required
- Chain rule
- Product rule
- Quotient rule
- Implicit differentiation
Questions may involve showing or proving given results or finding unknown constants
It is common that derivatives can be written in terms of the original function
- This is due to the derivative of e^x also being e^x giving rise to the repetition of terms

Use the same method for differentiating any inverse function
STEP 1
Write x in terms of y
- $y = arsinh x$ can be written $x = \sinh y$
STEP 2
Differentiate with respect to y
- $\frac{d x}{d y} = \cosh y$
STEP 3
Write the derivative in terms of x
- $\frac{d x}{d y} = \cosh y = \pm \sqrt{\sinh^{2} y + 1} = \pm \sqrt{x^{2} + 1}$
STEP 4
Take the reciprocal
- $\frac{d y}{d x} = \pm \frac{1}{\sqrt{x^{2} + 1}}$
STEP 5
Use the graph to determine whether it is positive of negative
- The graph of $y = arsinh x$ has a positive gradient everywhere
- $\frac{d y}{d x} = \frac{1}{\sqrt{x^{2} + 1}}$

It is usually easier to differentiate hyperbolic functions using the “trig style” standard results but if you are stuck you can try using their exponential form from the definitions

Given that

y = 2 x \sinh x

, show that

\frac{d^{2} y}{d x^{2}} = y + 4 \cosh x

RaTp7qLC_al-fm-4-1-4-we-solution-differentiation-a

Given that

f (x) = 3 arcosh 4 x

, show that

f' (5) = \frac{a}{\sqrt{b}}

where

a \in ℚ

and

b \in ℕ

are constants to be found.

al-fm-4-1-4-we-solution-differentiation-b

These are the reverse results of the derivatives, remembering “+c” of course!
- $\int \sinh x d x = \cosh x + c$
- $\int \cosh x d x = \sinh x + c$
  - These are given in the integration section of the formulae booklet
- $\int {sech}^{2} x d x = \tanh x + c$
  - This can be deduced from the differentiation section of the formulae booklet
There is also the integral of tanhx
- $\int \tanh x d x = \ln \cosh x + c$
  - This is given in the integration section of the formulae booklet
  - It can be shown using the substitution $u = \cosh x$
- Since cosh ≥ 1 for all values of x so there is no need for the modulus signs that usually accompany integrals involving ln

The following integration skills may be required
- Definite integration, area under a curve
- Reverse chain rule (‘adjust’ and ‘compensate’)
- Substitution
- Integration by parts
Hyperbolic identities may be required to rewrite an expression into an integrable form
For products involving e^x and a hyperbolic function use the definition involving e^x and e^-x for the hyperbolic function to write everything in terms of exponentials
- $\int e^{x} \cosh x d x = \frac{1}{2} \int e^{x} (e^{x} + e^{- x}) d x = \frac{1}{2} \int e^{2 x} + 1 d x$

To integrate inverse hyperbolic functions you would use integration by parts using the same technique as integrating lnx
- Write the functions as a product with 1
  - e.g. $arsinh x = 1 \times arsinh x$
- Differentiate the inverse function and integrate 1 when integrating by parts
  - $\int arsinh x d x = x arsinh x - \int \frac{x}{\sqrt{x^{2} + 1}} d x = x arsinh x - \sqrt{x^{2} + 1} + c$

Be aware of what is given in the formula booklet
- Practise using it to find integrals
- The results for hyperbolic functions and the inverse circular trig functions are listed together so try not to get confused
If you can't spot a relevant hyperbolic identity then using exponentials can make the expression easier and quicker to integrate