Derivatives Of Hyperbolic Function

Derivatives of hyperbolic functions

The derivatives of $\cosh x$ and $\sinh x$ are derived from their definitions:

Given $\sinh x = \frac{e^x - e^{-x}}{2}$, differentiating with respect to $x$ yields:

$\frac{d}{dx}(\sinh x) = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} = \cosh x$

Therefore, $\frac{d}{dx}(\sinh x) = \cosh x$.

Similarly, given $\cosh x = \frac{e^x + e^{-x}}{2}$, differentiating with respect to $x$ yields:

$\frac{d}{dx}(\cosh x) = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x + (-e^{-x})}{2} = \frac{e^x - e^{-x}}{2} = \sinh x$

Therefore, $\frac{d}{dx}(\cosh x) = \sinh x$.

Derivatives of other hyperbolic functions:

$\frac{d}{dx}(\tanh x) = \text{sech}^2 x$

$\frac{d}{dx}(\coth x) = -\text{cosech}^2 x$

$\frac{d}{dx}(\text{sech } x) = -\text{sech } x \tanh x$

$\frac{d}{dx}(\text{cosech } x) = -\text{cosech } x \coth x$

Example 1

Find the derivative of $y = \sinh(4x - 8)$ with respect to $x$.

Solution:

$\frac{dy}{dx} = \cosh(4x - 8) \cdot \frac{d}{dx}(4x - 8) = 4\cosh(4x - 8)$

Example 2

Find the derivative of $y = \ln(\tanh 2x)$ with respect to $x$.

Solution:

$\frac{dy}{dx} = \frac{1}{\tanh 2x} \cdot \frac{d}{dx}(\tanh 2x) = \frac{1}{\tanh 2x} \cdot \text{sech}^2 2x \cdot 2 = \frac{2\text{sech}^2 2x}{\tanh 2x} = 2 \frac{1}{\cosh^2 2x} \cdot \frac{\cosh 2x}{\sinh 2x} = \frac{2}{\sinh 2x \cosh 2x} = \frac{4}{2\sinh 2x \cosh 2x} = \frac{4}{\sinh 4x} = 4\text{cosech }4x$

Example 3

Find the minimum value of the function $y = 5\cosh x + 3\sinh x$.

Solution:

$\frac{dy}{dx} = 5\sinh x + 3\cosh x$

For a minimum, $\frac{dy}{dx} = 0$, so $5\sinh x + 3\cosh x = 0$.

$5\left(\frac{e^x - e^{-x}}{2}\right) + 3\left(\frac{e^x + e^{-x}}{2}\right) = 0$

$5e^x - 5e^{-x} + 3e^x + 3e^{-x} = 0$

$8e^x - 2e^{-x} = 0$

$8e^x = 2e^{-x}$

$4e^x = e^{-x}$

$e^{2x} = \frac{1}{4}$

$2x = \ln\left(\frac{1}{4}\right) = -\ln 4$

$x = -\frac{1}{2}\ln 4 = -\ln 2$

Substituting $x = -\ln 2$ back into the original equation:

$y = 5\cosh(-\ln 2) + 3\sinh(-\ln 2) = 5\left(\frac{e^{-\ln 2} + e^{\ln 2}}{2}\right) + 3\left(\frac{e^{-\ln 2} - e^{\ln 2}}{2}\right) = 5\left(\frac{1/2 + 2}{2}\right) + 3\left(\frac{1/2 - 2}{2}\right) = 5\left(\frac{5}{4}\right) + 3\left(\frac{-3}{4}\right) = \frac{25-9}{4} = \frac{16}{4} = 4$

Therefore, the minimum value is 4.

Example 4

Find the derivative of $y = \frac{e^{\cosh 6x} - 4\sinh 8x}{1 + \tanh 4x}$ with respect to $x$.

Solution:

Using the quotient rule:

$\frac{dy}{dx} = \frac{(1+\tanh 4x)\frac{d}{dx}(e^{\cosh 6x} - 4\sinh 8x) - (e^{\cosh 6x} - 4\sinh 8x)\frac{d}{dx}(1+\tanh 4x)}{(1+\tanh 4x)^2}$

$= \frac{(1+\tanh 4x)(e^{\cosh 6x} \cdot 6\sinh 6x - 32\cosh 8x) - (e^{\cosh 6x} - 4\sinh 8x)(4\text{sech}^2 4x)}{(1+\tanh 4x)^2}$

Derivatives of inverse hyperbolic functions

The derivative of inverse hyperbolic functions, such as $\sinh^{-1} x$, can be found using implicit differentiation.

For example, to find the derivative of $y = \sinh^{-1} x$:

$x = \sinh y$

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(x) = \frac{d}{dx}(\sinh y)$

$1 = \cosh y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{\cosh y}$

Using the identity $\cosh^2 y - \sinh^2 y = 1$, we have $\cosh y = \sqrt{1 + \sinh^2 y}$. Since $\sinh y = x$, we get $\cosh y = \sqrt{1 + x^2}$.

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

Therefore, $\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$.

Example 1

Find the derivative of $y = \cosh^{-1} x$ with respect to $x$.

Solution:

Let $y = \cosh^{-1} x$.

$x = \cosh y$

Differentiating both sides with respect to $x$:

$1 = \sinh y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{\sinh y}$

Using the identity $\cosh^2 y - \sinh^2 y = 1$, we have $\sinh y = \pm\sqrt{\cosh^2 y - 1} = \pm\sqrt{x^2 - 1}$. Since the range of $\cosh^{-1}$ is $y \ge 0$, we consider the positive root:

$\frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 1}}, \text{ for } x > 1$

Example 2

Find the derivative of each of the following functions with respect to $x$:

(a) $y = \sinh^{-1}\left(\frac{x}{3}\right)$

(b) $y = \ln(\cosh^{-1} x)$

Solution:

(a) $y = \sinh^{-1}\left(\frac{x}{3}\right)$

$\frac{dy}{dx} = \frac{1}{\sqrt{1 + (\frac{x}{3})^2}} \cdot \frac{1}{3} = \frac{1}{3\sqrt{1 + \frac{x^2}{9}}} = \frac{1}{\sqrt{9 + x^2}}$

(b) $y = \ln(\cosh^{-1} x)$

$\frac{dy}{dx} = \frac{1}{\cosh^{-1} x} \cdot \frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\cosh^{-1} x} \cdot \frac{1}{\sqrt{x^2 - 1}} = \frac{1}{\sqrt{x^2 - 1} \cosh^{-1} x}$

Power series expansions of cosh x and sinh x

The hyperbolic cosine and sine functions can be represented using Maclaurin series. The Maclaurin series for a function $f(x)$ is given by:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$

Power series expansion of $\cosh x$

The power series expansion for $\cosh x$ is:

$\cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$

Proof:

Let $y = \cosh x$.

$y(0) = \cosh 0 = 1$

$y'(x) = \sinh x \Rightarrow y'(0) = \sinh 0 = 0$

$y''(x) = \cosh x \Rightarrow y''(0) = \cosh 0 = 1$

$y'''(x) = \sinh x \Rightarrow y'''(0) = \sinh 0 = 0$

$y^{(4)}(x) = \cosh x \Rightarrow y^{(4)}(0) = \cosh 0 = 1$

And so on. Substituting these values into the Maclaurin series:

$\cosh x = 1 + 0x + \frac{1}{2!}x^2 + \frac{0}{3!}x^3 + \frac{1}{4!}x^4 + \dots$

$\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots$

Power series expansion of $\sinh x$

The power series expansion for $\sinh x$ is:

$\sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$

Proof:

Let $y = \sinh x$.

$y(0) = \sinh 0 = 0$

$y'(x) = \cosh x \Rightarrow y'(0) = \cosh 0 = 1$

$y''(x) = \sinh x \Rightarrow y''(0) = \sinh 0 = 0$

$y'''(x) = \cosh x \Rightarrow y'''(0) = \cosh 0 = 1$

And so on. Substituting these values into the Maclaurin series:

$\sinh x = 0 + 1x + \frac{0}{2!}x^2 + \frac{1}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5 + \dots$

$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots$

Example 1

Using the power series expansion of $\cosh x$ up to the term in $x^4$, approximate the value for $\cosh 1$, correct to 4 decimal places.

Solution:

$\cosh x \approx 1 + \frac{x^2}{2!} + \frac{x^4}{4!}$

$\cosh 1 \approx 1 + \frac{1^2}{2} + \frac{1^4}{24} = 1 + 0.5 + 0.041666... \approx 1.5417$

Example 2

Expand the expression $\sinh 3x$ as a power series as far as the term in $x^5$.

Solution:

$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots$

$\sinh 3x = 3x + \frac{(3x)^3}{3!} + \frac{(3x)^5}{5!} + \dots$

$\sinh 3x = 3x + \frac{27x^3}{6} + \frac{243x^5}{120} + \dots$

$\sinh 3x = 3x + \frac{9}{2}x^3 + \frac{81}{40}x^5 + \dots$

Example 3

Determine the power series for $\frac{1}{4}\cosh\frac{x}{4} - \sinh 2x$ as far as the term in $x^6$.

Solution:

$\frac{1}{4}\cosh\frac{x}{4} = \frac{1}{4}\left(1 + \frac{(\frac{x}{4})^2}{2!} + \frac{(\frac{x}{4})^4}{4!} + \frac{(\frac{x}{4})^6}{6!} + \dots\right) = \frac{1}{4} + \frac{x^2}{128} + \frac{x^4}{6144} + \frac{x^6}{786432} + \dots$

$\sinh 2x = 2x + \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} + \frac{(2x)^6}{6!} + ... = 2x + \frac{8x^3}{6} + \frac{32x^5}{120} + \frac{64x^6}{720} + ... = 2x + \frac{4}{3}x^3 + \frac{4}{15}x^5 + \frac{4}{45}x^6 + ...$

$\frac{1}{4}\cosh\frac{x}{4} - \sinh 2x = \left(\frac{1}{4} + \frac{x^2}{128} + \frac{x^4}{6144} + \frac{x^6}{786432} + \dots\right) - \left(2x + \frac{4}{3}x^3 + \frac{4}{15}x^5 + \frac{4}{45}x^6 + ...\right)$

$= \frac{1}{4} - 2x + \frac{1}{128}x^2 - \frac{4}{3}x^3 + \frac{1}{6144}x^4 - \frac{4}{15}x^5 + \left(\frac{1}{786432}-\frac{4}{45}\right)x^6 + ...$