Tutorial 8: Hyperbolic Functions

Hyperbolic Functions

1. Given sinh $x = \frac{5}{12}$, find:

  • a. $\cosh x$

    • Using $\cosh^2 x - \sinh^2 x = 1$: $$\cosh^2 x = 1 + \left(\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}$$ $$\cosh x = \frac{13}{12}$$

  • b. $\csc h , x$

    • $\csc h , x = \frac{1}{\sinh x} = \frac{1}{5/12} = \frac{12}{5}$

  • c. $\tanh x$

    • $\tanh x = \frac{\sinh x}{\cosh x} = \frac{5/12}{13/12} = \frac{5}{13}$

  • d. $\sinh 2x$

    • Using $\sinh 2x = 2 \sinh x \cosh x$: $$\sinh 2x = 2 \cdot \frac{5}{12} \cdot \frac{13}{12} = \frac{130}{144} = \frac{65}{72}$$

  • e. $\cosh 5x$

    • Use $\cosh 5x = \cosh(2x + 3x)$ or multiple angle formula; values can be derived from $\cosh x$ and $\sinh x$ using identities.

2. Show that $x = \frac{1}{2} \ln 3$ if $\tanh x = \frac{1}{2}$

  • Using $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$: $$\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{1}{2}$$ $$2(e^x - e^{-x}) = e^x + e^{-x}$$ $$2e^x - 2e^{-x} = e^x + e^{-x}$$ $$e^x = 3e^{-x}$$ $$e^{2x} = 3$$ $$x = \frac{1}{2} \ln 3$$

3. Find all possible values of $\sinh x$ for which $12 \cosh^2 x + 7 \sinh x = 24$

  • Use $\cosh^2 x = 1 + \sinh^2 x$: $$12(1 + \sinh^2 x) + 7 \sinh x = 24$$ $$12 \sinh^2 x + 7 \sinh x - 12 = 0$$
  • Let $u = \sinh x$: $$12u^2 + 7u - 12 = 0$$ $$(3u + 4)(4u - 3) = 0$$ $$u = -\frac{4}{3} \quad \text{or} \quad u = \frac{3}{4}$$
  • So $\sinh x = -\frac{4}{3}$ or $\frac{3}{4}$.

4. Find $\frac{dy}{dx}$ for:

  • a. $y = \sinh(4x - 8)$ $$\frac{dy}{dx} = 4 \cosh(4x - 8)$$

  • b. $y = \cosh(x^4)$ $$\frac{dy}{dx} = 4x^3 \sinh(x^4)$$

  • c. $y = \coth(\ln x)$

    • $\coth(\ln x) = \frac{\cosh(\ln x)}{\sinh(\ln x)} = \frac{x + 1/x}{x - 1/x} = \frac{x^2 + 1}{x^2 - 1}$
    • Differentiate: $$\frac{dy}{dx} = \frac{2x(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}$$

  • d. $y = \ln(\tanh 2x)$

    • Using chain rule: $$\frac{dy}{dx} = \frac{1}{\tanh 2x} \cdot \text{sech}^2 2x \cdot 2 = \frac{2 \text{sech}^2 2x}{\tanh 2x}$$

  • e. $y = \sinh^3(2x)$

    • Using chain rule: $$\frac{dy}{dx} = 3 \sinh^2(2x) \cdot \cosh(2x) \cdot 2 = 6 \sinh^2(2x) \cosh(2x)$$

5. Differentiate with respect to $x$:

  • a. $y = x^4 \cosh(3x^2 - 8x)$

    • Product rule: $$\frac{dy}{dx} = 4x^3 \cosh(3x^2 - 8x) + x^4 \sinh(3x^2 - 8x) \cdot (6x - 8)$$

  • b. $y = \cosh(\sin^{-1} x^4)$

    • Chain rule: $$\frac{dy}{dx} = \sinh(\sin^{-1} x^4) \cdot \frac{1}{\sqrt{1 - (x^4)^2}} \cdot 4x^3$$

  • c. $y = \coth(\ln x) \cos^{-1}(2x^3)$

    • Use result from 4c for $\coth(\ln x)$ derivative and product rule: $$\frac{dy}{dx} = \frac{-4x}{(x^2 - 1)^2} \cos^{-1}(2x^3) + \frac{x^2 + 1}{x^2 - 1} \cdot \frac{-1}{\sqrt{1 - 4x^6}} \cdot 6x^2$$

  • d. $y = \frac{\ln(\sin 2x)}{\coth(x^2)}$

    • Quotient rule: $$\frac{dy}{dx} = \frac{\frac{2 \cos 2x}{\sin 2x} \cdot \coth(x^2) - \ln(\sin 2x) \cdot (-\text{csch}^2(x^2) \cdot 2x)}{\coth^2(x^2)}$$

  • e. $y = \frac{\sinh^3(2x)}{\cosh^2(5x - 3) + x}$

    • Quotient rule with chain rule in numerator and denominator.

6. Evaluate integrals:

  • a. $\int \sinh^6 x \cosh x , dx$

    • Let $u = \sinh x$, $du = \cosh x , dx$: $$\int u^6 , du = \frac{u^7}{7} + C = \frac{\sinh^7 x}{7} + C$$

  • b. $\int \cosh(2x - 3) , dx$ $$\int \cosh(2x - 3) , dx = \frac{1}{2} \sinh(2x - 3) + C$$

7. Exact values:

  • a. $\tanh^{-1} 0 = 0$
  • b. $\cosh^{-1} \pi = \ln(\pi + \sqrt{\pi^2 - 1})$
  • c. $\sinh^{-1} \frac{\pi}{4} = \ln\left(\frac{\pi}{4} + \sqrt{1 + \frac{\pi^2}{16}}\right)$

8. Prove $\sinh^{-1} x = \ln(x + \sqrt{1 + x^2})$

  • Let $y = \sinh^{-1} x$, then $\sinh y = x$.
  • $\sinh y = \frac{e^y - e^{-y}}{2} = x$
  • Multiply by $2e^y$: $$e^{2y} - 1 = 2x e^y$$ $$e^{2y} - 2x e^y - 1 = 0$$
  • Solve quadratic in $e^y$: $$e^y = x \pm \sqrt{x^2 + 1}$$
  • Since $e^y > 0$, take positive root: $$e^y = x + \sqrt{x^2 + 1}$$
  • Thus: $$y = \ln(x + \sqrt{x^2 + 1})$$

9. Find $x$:

  • a. $\sinh^{-1} \frac{3}{4} + \sinh^{-1} x = \sinh^{-1}

Source: FAC1004 Tutorial 8 25-26.pdf