Hyperbolic Functions Revision Notes for Edexcel A-Level Further Mathematics

4.1.3 Differentiating & Integrating Hyperbolic Functions

Differentiating Inverse Hyperbolic and Trig Functions

Introduction to Inverse Hyperbolic Functions

Inverse hyperbolic functions, denoted as $\sinh^{-1} x, \cosh^{-1} x$ , and $\tanh^{-1} x$ , are the inverses of the hyperbolic sine, cosine, and tangent functions. These functions can be expressed in terms of natural logarithms, which are useful for simplifying and solving problems involving hyperbolic functions.

The logarithmic forms of inverse hyperbolic functions are derived using the definitions of hyperbolic functions and the properties of exponentials. These forms allow us to compute values for inverse hyperbolic functions without relying on numerical estimation of hyperbolic equations.

Logarithmic Forms of Inverse Hyperbolic Functions

Inverse Hyperbolic Sine ( $\sinh^{-1} x$ ):

\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}), \quad x \in \mathbb{R}

Inverse Hyperbolic Cosine ( $\cosh^{-1} x$ ):

\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}), \quad x \geq 1

Inverse Hyperbolic Tangent ( $\tanh^{-1} x$ ):

\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right), \quad |x| < 1

Worked Examples

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Example Differentiate $\arccos(x)$

Let $y = \arccos(x)$ and rearrange it to $x = f(y)$

y = \arccos(x) \implies x = \cos(y)

Find $\frac{dx}{dy}$

\frac{dx}{dy} = -\sin(y)

Find the reciprocal to get $\frac{dy}{dx}$ , then rearrange in terms of $x$

\frac{dy}{dx} = \frac{-1}{\sin(y)}

Given $x = \cos(y)$

x^2 = \cos^2(y) \implies 1 - \sin^2(y)

\sin^2(y) = 1 - x^2

\sin(y) = \pm\sqrt{1 - x^2}

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

(Redundant $\pm$ gives $-1$ in the denominator)

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

However, looking at the graph of $y = \arccos(x)$ :

At points ( $0, \pi/2$ ), ( $1,0$ ), and ( $-1, \pi$ ), the graph shows that the gradient is always negative.

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

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Example: Differentiate $y = \arccos(\ln(3x + 2))$

[Hint: Use the chain rule and the above result.]

\frac{dy}{dx} = \frac{-1}{\sqrt{1 - (\ln(3x + 2))^2}} \times \frac{1}{3x + 2} \times 3

= \frac{-3}{(3x + 2)\sqrt{1 - [\ln(3x + 2)]^2}}

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Example: Differentiate $y = \arccos(e^x + 7x)$

\frac{dy}{dx} = \frac{-1}{\sqrt{1 - (e^x + 7x)^2}} \times (e^x + 7)

= \frac{-(e^x + 7)}{\sqrt{1 - (e^x + 7x)^2}}

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Example: Find in full, showing detailed reasoning, the first derivative of $\arcsin(x)$

y = \arcsin(x) \implies x = \sin(y)

\frac{dx}{dy} = \cos(y) \implies \frac{dy}{dx} = \frac{1}{\cos(y)}

Since $x = \sin(y)$ ,

x^2 = \sin^2(y) = 1 - \cos^2(y)

\implies \cos^2(y) = 1 - x^2 \implies \cos(y) = \pm\sqrt{1 - x^2}

Thus:

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

Note that the gradient is always positive, so:

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

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Example: Differentiate $y = \text{arsinh}(x)$

\sinh(y) = x \implies \frac{dx}{dy} = \cosh(y) \implies \frac{dy}{dx} = \frac{1}{\cosh(y)}

Since $\cosh^2(y) - \sinh^2(y) = 1$

\cosh^2(y) = 1 + \sinh^2(y) \implies \cosh(y) = \pm \sqrt{1 + \sinh^2(y)}

\therefore \frac{dy}{dx} = \pm\frac{1}{\sqrt{1 + sinh^2(y)^2}} = \frac{dy}{dx} = \pm\frac{1}{\sqrt{1 + x^2}}

Since gradient is always Positive

\therefore \quad \frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}

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Example: Use the above result to find $\frac{d}{dx} (\text{arsinh}(\tan{x}))$

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

Let $y = \text{arsinh}(\tan{x})$ , then

\frac{dy}{dx} = \frac{1}{\sqrt{1 + \tan^2{x}}} \times \sec^2{x}

Since $1 + \tan^2{x} = \sec^2{x}$ , we get:

\frac{dy}{dx} = \frac{\sec^2{x}}{\sqrt{\sec^2{x}}} = \sec{x}

Past Paper Worked Questions

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Q1. (June 2006, Q2) (i) Given that $y = \tan^{-1}(x)$ , prove that $\frac{dy}{dx} = \frac{1}{1 + x^2}$ .

(ii) Verify that $y = \tan^{-1}(x)$ satisfies the equation

(1 + x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = 0.

Solution:

(i) Let $y = \arctan(x)$ .

\implies x = \tan(y) \implies \frac{dx}{dy} = \sec^2(y)

Thus,

\frac{dy}{dx} = \frac{1}{\sec^2(y)} = \frac{1}{\tan^2(y) + 1} = \frac{1}{1 + x^2}

(ii) $\frac{dy}{dx} = (1 + x^2)^{-1}$

Differentiate again:

\frac{d^2y}{dx^2} =-(1 + x^2)^{-2} = -\frac{2x}{(1 + x^2)^2}

Substitute $\frac{dy}{dx}\ and \ \frac{d^2y}{dx^2}$ into the LHS of the expression:

\text{LHS} = (1 + x^2) \times \left( -\frac{2x}{(1 + x^2)^2} \right) + 2x \left( \frac{1}{1 + x^2} \right)

Simplify:

= \frac{-2x\color {red}\cancel{(1 + x^2)}}{(1 + x^2) \color {red} \cancel{^2}} + \frac{2x}{1 + x^2}

= \frac{-2x}{1 + x^2} + \frac{2x}{1 + x^2} = 0

Thus, the equation is verified.

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Q4. (Jan 2010, Q9) (i) Given that $y = \tanh^{-1}(x)$ , for $-1 < x < 1$ , prove that $y = \frac{1}{2} \ln \left( \frac{1 + x}{1 - x} \right)$ .

Let $x = \tanh(y) = \frac{\sinh(y)}{\cosh(y)}$ .

x = \tanh y = \frac{\sinh y}{\cosh y} = \frac{\frac{1}{2}(e^y - e^{-y})}{\frac{1}{2}(e^y + e^{-y})} = \frac{(e^y - e^{-y})}{(e^y + e^{-y})}

\Rightarrow x(e^y + e^{-y}) = e^y - e^{-y}

Multiply by $e^y$

\Rightarrow x(e^{2y} + 1) = e^{2y} - 1 \Rightarrow xe^{2y} + x = e^{2y} - 1

\Rightarrow xe^{2y} - e^{2y} = -1 - x

\Rightarrow e^{2y}(x - 1) = -1 - x \Rightarrow e^{2y} = \frac{-1 - x}{x - 1} \times \frac {-1}{-1} = \frac{1 + x}{1 - x}

\Rightarrow 2y = \ln \left( \frac{1+x}{1-x} \right) \Rightarrow y = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) \quad \text{as required}

(ii) Given that $f(x) = a \cosh(x) - b \sinh(x)$ , where a and b are positive constants:

(a) Given that $b \geq a$ , show that the curve with equation $y = f(x)$ has no stationary points.

f^1(x) = a \sinh(x) - b \cosh(x)=0 \implies a\sinh (x) = b \cosh (x)

\implies \frac {\sinh(x)}{\cosh(x)} = \frac{b}{a} \implies \tanh (x) = \frac{b}{a}

\implies x = archtanh \bigg (\frac{b}{a} \bigg)

Since $\frac{b}{a} \geq 1$ and $archtanh(x)$ has domain $-1 < x < 1$

$\therefore$ no solutions and no stationary points

(b) In the case where $a > 1$ and $b = 1$ , show that $f(x)$ has a minimum value of $\sqrt{a^2 - 1}$ .

Let $f(x) = a \cosh(x) - \sinh(x)$ .

Differentiate:

f'(x) = a \sinh(x) - \cosh(x) = 0\\ a \sinh (x) = \cosh (x) \implies \tanh (x) = \frac {1}{a}

\implies \sinh(x) = \frac{1}{a} \cosh(x) \implies \tanh(x) = \frac{1}{a}

x = ar\tanh \left( \frac{1}{a} \right) = \frac{1}{2} \ln \left( \frac{1 + \frac{1}{a}}{1 - \frac{1}{a}} \right) = \frac{1}{2} \ln \left( \frac{a + 1}{a - 1} \right) = \ln \left(\sqrt \frac{a + 1}{a - 1} \right)

f(x) =a \cosh (x)-\sinh (x)= \frac{a}{2} \left( e^{\ln \left( \frac{a + 1}{a - 1} \right)} + e^{-\ln \left( \frac{a + 1}{a - 1} \right)} \right) - \frac{1}{2} \left( e^{\ln \left( \frac{a + 1}{a - 1} \right)} - e^{-\ln \left( \frac{a + 1}{a - 1} \right)} \right)

Simplifying:

= \frac{a}{2} \left( \sqrt \frac{a + 1}{a - 1} + \sqrt\frac{a - 1}{a + 1} \right) - \frac{1}{2} \left( \sqrt\frac{a + 1}{a - 1} - \sqrt\frac{a - 1}{a + 1} \right)

Finally, this simplifies to $\sqrt{a^2 - 1}$ .

= \frac{a}{2} \left( \frac{ \sqrt{(a+1)^2} + \sqrt{(a-1)^2}}{\sqrt{(a-1)(a-1)}} \right) - \frac{1}{2} \left( \frac{ \sqrt{(a+1)^2} - \sqrt{(a-1)^2}}{\sqrt{(a+1)(a-1)}} \right)

= \frac{a}{2} \left( \frac{a\cancel{+1} + a\cancel{-1}}{\sqrt{(a+1)(a-1)}} \right) - \frac{1}{2} \left( \frac{a+1 - a + 1}{\sqrt{(a+1)(a-1)}} \right)

= \frac{a}{\cancel2} \left( \frac{\cancel2a}{\sqrt{a^2 - 1}} \right) - \frac{1}{\cancel2} \left( \frac{\cancel2}{\sqrt{a^2 - 1}} \right)

= \frac{a^2}{\sqrt{a^2 - 1}} - \frac{1}{\sqrt{a^2 - 1}} = \frac{a^2 - 1}{\sqrt{a^2 - 1}} = \sqrt{a^2 - 1} \quad \text{as required}

Note Summary

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Common Mistakes:

Confusing the formulas for $\sinh^{-1} x$ and $\cosh^{-1} x$ : For $\sinh^{-1} x$ , the square root is $\sqrt{x^2 + 1}$ ; for $\cosh^{-1} x$ , it is $\sqrt{x^2 - 1}$
Using incorrect domains: $\cosh^{-1} x$ is only valid for $x \geq 1$ , and $\tanh^{-1} x$ is only valid for $|x| < 1$
Sign errors when solving quadratic equations: Ensure you select the correct branch of the logarithmic solution based on the context.
Overlooking restrictions on $\ln$ : Logarithmic functions require positive arguments, so ensure $x + \sqrt{x^2 + 1} > 0$

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Key Formulas:

$\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}) x \in \mathbb{R}$
$\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}) x \geq 1$
$\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right) |x| < 1$
Domains:

$\sinh^{-1} x: x \in \mathbb{R}$
$\cosh^{-1} x: x \geq 1$
$\tanh^{-1} x: |x| < 1$

Hyperbolic Functions (Edexcel A-Level Further Mathematics): Revision Notes

4.1.3 Differentiating & Integrating Hyperbolic Functions

Differentiating Inverse Hyperbolic and Trig Functions

Introduction to Inverse Hyperbolic Functions