Derivatives of Hyperbolic Functions

(Video) Derivatives of Hyperbolic Functions

by PatrickJMT

Before we begin we should preface this with hyperbolic functions suck. Show me someone who likes hyperbolic functions and I'll show you a liar. (These are my thoughts on the subject, not Patrick's.)

Hyperbolic functions are functions that determine the right half of an equilateral hyperbola. Just like trigonometric functions determine a circle.

Recall the definitions of the hyperbolic functions: $$\begin{alignat}{2} & \sinh x =\frac{e^x-e^{-x}}{2} \quad && \text{csch } x=\frac{1}{\sinh x}\\ & \cosh x=\frac{e^x+e^{-x}}{2} \quad && \text{sech }x=\frac{1}{\cosh x}\\ & \tanh x=\frac{\sinh x}{\cosh x} \quad && \text{coth }x=\frac{\cosh x}{\sinh x} \end{alignat}$$

The derivatives of the hyperbolic functions are as follows: $$\begin{align} \frac{d}{dx}\sinh x&=\cosh x\\ \frac{d}{dx}\cosh x&=\sinh x\\ \frac{d}{dx}\tanh x&=\text{sech}^2\ x\\ \frac{d}{dx}\text{csch }x&=-\text{csch }x\ \text{coth }x\\ \frac{d}{dx}\text{sech }x&=-\text{sech }x\ \text{tanh }x\\ \frac{d}{dx}\text{coth }x&=-\text{csch}^2\ x \end{align}$$ Notice that the first three are positive and the final three are negative. Besides that, the derivatives are pretty much the same as the derivatives of the trig functions.

To see how these derivatives are derived, we just need to use their respective definitions. For example, let's prove $\frac{d}{dx}\sinh x=\cosh x$. $$\begin{align} \frac{d}{dx}\sinh x&=\frac{d}{dx}\bigg(\frac{e^x-e^{-x}}{2}\bigg)\\ &=\frac{d}{dx}\bigg[\frac{e^x}{2}-\frac{e^{-x}}{2}\bigg]\\ &=\frac{e^x}2-\bigg(-\frac{e^{-x}}{2}\bigg)\\ &=\frac{e^x+e^{-x}}{2}\\ &=\boxed{\cosh x} \end{align}$$

Finding the derivative of hyperbolic functions is as standard as other functions. You just need to remember your chain rule, product rule, and quotient rules really. To demonstrate this he looks at some examples.

The first problem he looks at is finding the derivative of: $$f(x)=\tanh(4x)$$ All we have to do here is use our chain rule. $$\begin{align} f'(x)&=\text{sech}^2(4x)\cdot\color{red}{[4x]'}\\ &=\text{sech}^2(4x)\cdot4\\ &=\boxed{4\text{sech}^2(4x)} \end{align}$$

The next problem he looks at is: $$\begin{align} f(x)&=\ln(\sinh x)\\ f'(x)&=\frac{1}{\sinh x}\cdot\color{red}{[\sinh x]'}\\ &=\frac{1}{\sinh x}\cdot\cosh x\\ &=\frac{\cos h}{\sin h}\\ &=\boxed{\text{coth } x} \end{align}$$

The final example he solves is: $$\begin{align} f(x)&=\sinh x\ \tanh x\\ f'(x)&=\color{red}{[\sinh x]'}\tanh x+\sinh x\color{red}{[\tanh x]'}\\ &=\cosh x\ \tanh x+\sinh x\ \text{sech}^2\ x\\ &=\cosh x\cdot\frac{\sinh x}{\cosh x}+\sinh x\cdot\text{sech}^2\ x\\ &=\sinh x+\sinh x\cdot\text{sech}^2\ x\\ &=\boxed{\sinh x(1+\text{sech}^2\ x)} \end{align}$$

(Video) Derivatives of Inverse Hyperbolic Functions

by PatrickJMT

Here's a video by patrickJMT showing you how to take derivatives of inverse hyperbolic functions. These are similar to inverse trig functions and not much trickier to use.

The derivatives of inverse hyperbolic functions are as follows: $$\begin{align} \frac{d}{dx}\sinh^{-1}x&=\frac{1}{\sqrt{1+x^2}}\\ \frac{d}{dx}\cosh^{-1}x&=\frac{1}{\sqrt{x^2-1}}\\ \frac{d}{dx}\tanh^{-1}x&=\frac{1}{1-x^2};\ |x|<1\\ \frac{d}{dx}\text{csch}^{-1}\ x&=-\frac{1}{|x|\sqrt{x^2+1}}\\ \frac{d}{dx}\text{sech}^{-1}\ x&=-\frac{1}{x\sqrt{1-x^2}}\\ \frac{d}{dx}\text{coth}^{-1}\ x&=\frac{1}{1-x^2};\ |x|>1 \end{align}$$

The first example he looks is the function: $$y=x^2\sinh^{-1}(2x)$$ Nothing complicated here, we just need to use the product rule. $$\begin{align} y'&=\color{red}{[x^2]'}\sinh^{-1}(2x)+x^2\color{red}{[\sinh^{-1}(2x)]'}\\ &=2x\sinh^{-1}(2x)+x^2\cdot\frac{1}{\sqrt{1+(2x)^2}}\cdot\color{red}{[2x]'}\\ &=2x\sinh^{-1}(2x)+x^2\cdot\frac{1}{\sqrt{1+4x^2}}\cdot2\\ &=\boxed{2x\bigg(\sinh^{-1}(2x)+\frac{x}{\sqrt{1+4x^2}}\bigg)} \end{align}$$

The next example that he looks at is: $$y=\tanh^{-1}(\sqrt x)$$ As always it's good practice to rewrite the problem without any radical symbols. $$\begin{align} y&=\tanh^{-1}(x^{\frac12})\\ y'&=\frac{1}{1-(x^{\frac12})^2}\cdot\color{red}{[x^{\frac12}]'}\\ &=\frac{1}{1-(x^{\frac12})^2}\cdot\frac{1}{2\sqrt{x}}\\ &=\frac{1}{1-x}\cdot\frac{1}{2\sqrt{x}}\\ &=\boxed{\frac{1}{2\sqrt{x}(1-x)}} \end{align}$$

The final example he looks at is: $$\begin{align} y&=\text{sech}^{-1}\sqrt{1-x^2}\\ &=\text{sech}^{-1}[(1-x^2)^{\frac12}]\\ y'&=-\frac{1}{(1-x^2)^{\frac12}\sqrt{1-[(1-x^2)^{\frac12}]^2}}\cdot\color{red}{[(1-x^2)^{\frac12}]'}\\ &=-\frac{1}{(1-x^2)^{\frac12}\sqrt{1-[(1-x^2)^{\frac12}]^2}}\cdot\frac{1}{2\sqrt{1-x^2}}\cdot\color{red}{[1-x^2]'}\\ &=-\frac{1}{(1-x^2)^{\frac12}\sqrt{1-[(1-x^2)^{\frac12}]^2}}\cdot\frac{1}{2\sqrt{1-x^2}}\cdot(-2x)\\ &=\frac{1}{(1-x^2)^{\frac12}\sqrt{1-1+x^2}}\cdot\frac{x}{\sqrt{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}\sqrt{x^2}}\cdot\frac{x}{\sqrt{1-x^2}}\\ &=\frac{1}{x\sqrt{1-x^2}}\cdot\frac{x}{\sqrt{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}}\cdot\frac{1}{\sqrt{1-x^2}}\\ &=\boxed{\frac{1}{1-x^2}} \end{align}$$