\(\DeclareMathOperator{\sech}{sech}\DeclareMathOperator{\csch}{cosech}\DeclareMathOperator{\coth}{coth}\)Imagine you are on a skateboard, skating along a suspension bridge like the one pictured below. At any point, what is the gradient of the line pointing in the same direction as your skateboard? If you knew the shape of the line that the bridge maps out, this would be a simple task; just differentiate this line and plug in the point at which your skateboard is currently located. But what shape is the bridge? It just so happens that hyperbolic functions hold the answer: a suspension bridge is a catenary curve, the name of the curve created by the hyperbolic cosine function. Thus to figure out the direction of your skateboard, you need to know how to differentiate the hyperbolic cosine function.
A suspension bridge, just like any other curve created by letting something hang between two points naturally under gravity, is a catenary curve. This is the name of the curve mapped out by the hyperbolic cosine function. Hans Braxmeier, Pixabay.
Formulas for the Differentiation of Hyperbolic Functions
The hyperbolic trigonometric functions are similar to the normal trigonometric functions, but instead of mapping out the unit circle, they map out the unit hyperbola.
Fig. 1. Hyperbolic sine and cosine can be derived from the unit hyperbola (\(x^2 - y^2 = 1 \)) in a very similar way to how the standard sine and cosine functions can be derived from the unit circle.
The standard hyperbolic functions are:
- Hyperbolic sine: \( \sinh{x} \).
- Hyperbolic cosine: \( \cosh{x} \).
- Hyperbolic tangent: \( \tanh{x} \).
Similarly, the reciprocal hyperbolic functions are:
- Hyperbolic secant: \( \sech{x} \).
- Hyperbolic cosecant: \( \csch{x} \).
- Hyperbolic cotangent: \( \coth{x} \).
For more information about hyperbolic functions, including their exponential forms, see Hyperbolic Functions.
Formulas for the Derivatives of Hyperbolic Sine, Cosine and Tangent
The derivatives of the standard hyperbolic functions are:
\[\begin{align} \frac{d}{dx} \sinh{x} & = \cosh{x}, \\ \frac{d}{dx} \cosh{x} & = \sinh{x}, \\ \frac{d}{dx} \tanh{x} & = \sech^{2}{x}. \end{align} \]
Formulas for the Derivatives of Reciprocal Hyperbolic Functions
The derivatives of the reciprocal hyperbolic functions are:
\[ \begin{align} \frac{d}{dx} \sech{x}, & = - \sech{x} \tanh{x}, \\ \frac{d}{dx} \csch{x}, & = - \csch{x} \coth{x}, \\ \frac{d}{dx} \coth{x} & = - \csch^2{x}. \end{align}\]
You should notice how closely these resemble the derivatives of trigonometric functions, the key difference being whether the result is positive or negative when there is a hyperbolic sine involved.
To prove these derivatives, it is often useful to have the hyperbolic functions in exponential form.
Prove that the derivative of \( \cosh(x) \) is \( \sinh(x) \).
Answer:
Firstly, write \( \cosh(x) \) in exponential form.
\[ \cosh{x} = \frac{e^{x} + e^{-x}}{2} = \frac{e^{x}}{2} + \frac{e^{-x}}{2}.\]
Now, as you take the derivative, you can use the chain rule and the fact that \( \frac{d}{dx} e^x = e^x\):
\[ \begin{align} \frac{d}{dx} \cosh{x} & = \frac{e^{x}}{2} - \frac{e^{-x}}{2} \\ & = \frac{e^{x} - e^{-x}}{2} \\ & = \sinh{x}. \end{align} \]
You can use the same method to prove the derivative of \( \sinh{x} \) is \( \cosh{x} \). To prove the derivatives of the reciprocal hyperbolic functions, you can use the quotient rule.
Prove \(\frac{d}{dx} \sech{x} = - \sech{x} \tanh{x} \).
Answer:
Write \( \sech{x} \) in terms of \( \cosh{x} \).
\[ \frac{d}{dx} \sech{x} = \frac{d}{dx} \frac{1}{\cosh{x}}.\]
From here, you can use the quotient rule to get
\[ \begin{align} \frac{d}{dx} \sech{x} & = \frac{ \cosh{x} \frac{d}{dx} (1) - 1 \cdot \frac{d}{dx} (\cosh{x}) }{cosh^2{x}} \\ & = \frac{0- \sinh{x} }{\cosh^2{x}} \\ & = -\frac{\sinh{x}}{\cosh^2{x}}. \end{align} \]
Now, separate the fraction into the two parts, \( \sech{x} \) and \( \tanh{x} \):
\[ \begin{align} \frac{d}{dx} \sech{x} & = - \frac{\sinh{x}}{\cosh^2{x}} \\ & = - \left(\frac{1}{\cosh{x}}\right) \left(\frac{\sinh{x}}{\cosh{x}}\right) \\ & = - \sech{x} \tanh{x}, \end{align} \]
as required.
The method for proving the derivatives of \( \coth{x} \) and \( \csch{x} \) is the same.
Definition of Hyperbolic Functions Using Differential Equations
Now that you know how to differentiate hyperbolic functions, you can now consider another way of defining the hyperbolic functions. The trigonometric functions sine and cosine are the solution to the following system of differential equations:
- \( c'(x) = s(x) \),
- \( s'(x) = - c(x) \)
with the initial conditions \( c(0) = 1 \), \( s(0) = 0 \).
But now, consider the same system of differential equations with the minus sign removed. This is where the hyperbolic trigonometric functions come in. The hyperbolic functions are the solution to the system of differential equations:
- \( c'(x) = s(x) \),
- \( s'(x) = c(x) \)
with the same initial conditions as before.
Furthermore, a solution to the first order differential equation \( y' = y \) with initial condition \( y(0) = 1 \) is \( e^x \). But what about the similar second order differential equation \( y'' = y \), with initial conditions \( y(0) = 1 \), \( y'(0) = 0 \)?
Prove that \(\cosh{x}\) is a solution to the differential equation \( y'' = y \), with initial conditions \( y(0) = 0 \), \( y'(0) = 0 \).
Answer:
Set \( y = \cosh{x} \). You can differentiate this using the formulas you saw earlier, to get: \( y' = \sinh{x} \). If you differentiate this again, you will get \( y'' = \cosh{x} = y \), so it satisfies the second order differential equation.
All that is left to do is prove that the initial conditions hold true as well. Since:
\[ y(x) = \cosh{x} = \frac{e^x + e^{-x}}{2}, \] we can substitute \( x = 0 \) into this to get:
\[ y(0) = \frac{e^0 + e^{-0}}{2} = 1.\]
Finally, we can write the differential of \( y\) in exponential form: \[y'(x) = \sinh{x} = \frac{e^x - e^{-x}}{2}, \] and again we can substitute \( x = 0 \) into this to get:
\[ y'(0) = \frac{e^0 - e^{-0}}{2} = 0,\]
As required. This concludes the proof.
Note that hyperbolic sine is also a solution to the same differential equation, but with the initial conditions \( y(0) = 0\), \( y'(0) = 0 \).
What else can be done to a number, such that when it is done twice, the number will stay the same but with the opposite sign? This would be multiplying the number by the imaginary unit \( i \). For a recap on imaginary numbers, see Core Complex Numbers. By substituting \(x\) for \(ix\) into the equations, you get the following results:
- \( \cosh{x} = \cos{ix}, \)
- \( \sinh{x} = -i \sin{ix}. \)
From this, you can derive an exponential form for the trigonometric functions, just like the exponential form for the hyperbolic functions. These are:
\[ \begin{align} \cos{x} & = \frac{e^{ix} + e^{-ix}}{2}, \\ \sin{x} & = \frac{e^{ix} - e^{-ix}}{2i}. \end{align}\]
These formulas can also be found using Maclaurin series expansions. See Maclaurin Series.
Now if you multiply the second equation, \(\sin{x}\), by the imaginary unit \(i \), you will get:
\[ i \sin{x} = \frac{e^{ix} - e^{-ix}}{2}. \]
Finally, you can add the exponential formula for \( \cos{x}\) to the formula above for \(i \sin{x}\), to get:
\[ \begin{align} \cos{x} + i \sin{x} & = \frac{e^{ix} + e^{-ix}}{2} + \frac{e^{ix} - e^{-ix}}{2} \\ \implies e^{ix} & = \cos{x} + i \sin{x}. \end{align}\]
This equation is known as Euler's Formula. It is an incredibly famous result in mathematics. Substituting in \(x = \pi \) into this formula will yield the result known as Euler's Identity:
\[ e^{i \pi} = -1. \]
Examples of Differentiation of Hyperbolic Trigonometric Functions
Let's take a look at a question where you must find the differential of a hyperbolic function using the product and chain rule.
Find \( \frac{dy}{dx} \), where \( y = x^2 \sinh{e^x} \).
Answer:
Firstly, you can use the product rule to get:
\[ \begin{align} \frac{dy}{dx} & = \frac{d}{dx} (x^2) \sinh(e^x) + x^2 \frac{d}{dx} (\sinh{e^x}) \\ & = 2 x \sinh(e^x) + x^2 \frac{d}{dx} (\sinh{e^x}). \end{align} \]
All that is left now is to differentiate the hyperbolic sine part, \( \sinh{e^x} \). Using the chain rule: \( \frac{d}{dx} ( \sinh{e^x} ) = e^x \cosh{e^x} \). Thus, the final answer is: \[ \frac{dy}{dx} = 2 x \sinh{e^x} + x^{2} e^{x} \cosh{e^x}. \]
Another way of tackling questions with hyperbolic functions is to convert the hyperbolic function into exponential form first. Let's take a look at using the exponential form of a hyperbolic function to make taking the derivative simpler.
Differentiate \(f(x) = e^{x} \cosh{x} \). First, lets convert this to exponential form. \[ \begin{align} f(x) & = e^{x} \cosh{x} \\ & = e^{x} \left( \frac{e^x + e^{-x}}{2} \right) \\ & = \frac{e^{2x}}{2} + \frac{1}{2}. \end{align} \] This function is much simpler to differentiate. Taking the derivative, you can see that: \[ f'(x) = e^{2x}. \]
Formulas for Differentiation of Inverse Hyperbolic Functions
The derivatives of the inverse hyperbolic functions are:
\[ \begin{align} \frac{d}{dx} \sinh^{-1}{x} & = \frac{1}{\sqrt{1+x^2}}, \\ \frac{d}{dx} \cosh^{-1}{x} & = \frac{1}{\sqrt{x^2 - 1}}, \\ \frac{d}{dx} \tanh^{-1}{x} & = \frac{1}{1-x^2}. \end{align}\]
Again, you will notice a resemblance to the derivatives of the inverse trigonometric functions. Knowing all the inverse hyperbolic and trigonometric derivatives will make solving many complicated integrals much easier. See Inverse Hyperbolic Functions for more details.
Examples of Differentiation of Inverse Hyperbolic Functions
Common questions involving differentiating inverse hyperbolic functions might be similar to questions about differentiating standard hyperbolic functions. They could include the chain, product or quotient rules, similar to the ones you have seen in this article. Something else you might be asked to do is to prove similar formulas to the ones listed above. For example you might be asked to prove that
\[ \frac{d}{dx} \left( \frac{1}{a} \tanh^{-1}{bx} \right) = \frac{b}{a(1 - (bx)^2)}. \]
For more information on how to solve questions like these, see Inverse Hyperbolic Functions.
Differentiating Hyperbolic Functions - Key takeaways
- Differentiating hyperbolic functions works almost the same way as normal trigonometric functions, except that sometimes you will have get a different sign whenever there is a hyperbolic sine involved.
- You can use the hyperbolic function identities to find these derivatives, or write the hyperbolic functions in exponential formand then take the derivative.
- If you differentiate the basic hyperbolic functions, you get:\[ \begin{align} \frac{d}{dx} \sinh{x} & = \cosh{x}, \\ \frac{d}{dx} \cosh{x} & = \sinh{x}, \\ \frac{d}{dx} \tanh{x} & = \sech^{2}{x}. \end{align}\]
If you differentiate the reciprocal hyperbolic functions, you get:\[ \begin{align} \frac{d}{dx} \sech{x} & = - \sech{x} \tanh{x},\\ \frac{d}{dx} \csch{x} &= - \csch{x} \coth{x}, \\ \frac{d}{dx} \coth{x} & = - \csch^2{x}. \end{align}\]
If you differentiate the inverse hyperbolic functions, you get:
\[ \begin{align} \frac{d}{dx} \sinh^{-1}{x} & = \frac{1}{\sqrt{1+x^2}}, \\ \frac{d}{dx} \cosh^{-1}{x} & = \frac{1}{\sqrt{x^2 - 1}}, \\ \frac{d}{dx} \tanh^{-1}{x} & = \frac{1}{1-x^2}. \end{align} \]
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