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Derivatives of Hyperbolic Functions

Calculus 2 · Chapter 3: Hyperbolic FunctionsSection 3.3 · lesson 14 of 5411 min · free

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In this lesson

Differentiating the family built from the exponential.

  • Derive \sinh' and \cosh' straight from the definitions
  • Get the other four from the quotient rule
  • See exactly where the trigonometric sign pattern breaks
  • Apply the chain rule to real composites

Transcript

Everything said in the video, section by section (1,583 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 3.3

Derivatives of Hyperbolic Functions

A power line strung between two poles. A chain hanging from two hooks. The cables of a suspension bridge before the deck is attached. Each settles into the same graceful curve. And as we saw, that curve is not a parabola. It's a catenary, described by the hyperbolic cosine — y equals a cosh of x over a. So to find the slope of such a cable at any point, or the angle at which it meets its anchor, you have to differentiate a hyperbolic function. The same need arises in special relativity, where rapidity is measured with inverse hyperbolic functions. And in the analysis of certain electrical transmission lines. Differentiating these is a practical skill, not a curiosity. And here's the good news. Because every hyperbolic function is ultimately built from e to the x and e to the minus x — whose derivatives you already know — no new differentiation principle is needed. Only careful bookkeeping.

The Foundation

No New Principle - Only Careful Bookkeeping

Let's start from the definitions. Sinh x is e to the x minus e to the minus x, over two. Differentiating is a direct application of two facts you know. The derivative of e to the x is e to the x. And by the chain rule, the derivative of e to the minus x is minus e to the minus x. So differentiating term by term, you get e to the x minus, minus e to the minus x. All over two. The two minuses make a plus. So you get e to the x plus e to the minus x, over two. Which is exactly cosh x. Now differentiate cosh x. Same procedure. You get e to the x plus, minus e to the minus x, over two. Which is e to the x minus e to the minus x, over two. Exactly sinh x. These two results are the cornerstone of the entire section. And they reveal a structure worth pausing on. In ordinary trigonometry, differentiating cosine introduces a minus sign. The derivative of cosine x is minus sine x. In the hyperbolic world that minus is absent. Cosh and sinh differentiate into each other cleanly, with no sign flip at all. And that single difference accounts for most of the sign discrepancies between the trigonometric and hyperbolic derivative formulas.

Figure 3.3

Each Function Plotted With Its Own Derivative

Look at the picture, because it makes the relationship concrete. Each function is plotted together with its own derivative. On the left, sinh x in solid, and cosh x dashed. And the claim is that the slope of sinh at every point equals the height of cosh. Check it. At x equals zero, sinh has its gentlest slope — and cosh is at its minimum, one. So the slope there should be one. It is. Far to the right, sinh is climbing steeply, and cosh is large. Consistent. On the right panel, cosh solid and sinh dashed. The slope of cosh equals the height of sinh. Notice something nice. Cosh has its minimum exactly where sinh crosses zero. Which is precisely what it means for sinh to be its derivative. And in both panels — unlike the trigonometric case — neither derivative carries a negative sign anywhere.

The Remaining Four

Everything Else Comes From the Quotient Rule

The remaining four hyperbolic functions are defined as ratios and reciprocals of sinh and cosh. So their derivatives follow from the quotient rule, together with the two results we just established. Nothing new required. Start with tanh x, which is sinh over cosh. Apply the quotient rule. You get cosh times cosh, minus sinh times sinh, all over cosh squared. The numerator is cosh squared minus sinh squared. And by the fundamental identity from section three point two, that's exactly one. So the derivative is one over cosh squared. Which is sech squared x. And notice — that's the exact hyperbolic analogue of the derivative of tangent being secant squared. Here even the sign agrees. Now sech x, which is one over cosh x. Write it as cosh x to the minus one, and apply the chain rule. You get minus cosh to the minus two, times sinh x. Which simplifies to minus sech x tanh x. Similar computations complete the set. For coth x, the quotient rule numerator becomes minus the quantity cosh squared minus sinh squared. Which is minus one. So the derivative of coth is minus csch squared x. And the derivative of csch is minus csch x coth x. Now here's a pattern worth noticing. The three derivatives carrying a negative sign — those of sech, csch, and coth — are precisely the ones whose names begin with a reciprocal prefix. Which mirrors exactly the pattern from trigonometric derivatives. Nothing to memorize separately.

Table 3.3

All Six, Beside Their Trigonometric Counterparts

Here's the full set in one table, with the trigonometric counterpart alongside each one. The derivative of sinh is cosh. Trig analogue: the derivative of sine is cosine. No sign change. The derivative of cosh is sinh. Trig analogue: the derivative of cosine is minus sine. The sign differs. This is the exception. The derivative of tanh is sech squared. Trig: secant squared. Matches. The derivative of coth is minus csch squared. Trig: minus cosecant squared. Matches. The derivative of sech is minus sech tanh. Trig: minus secant tangent. Matches. And the derivative of csch is minus csch coth. Trig: minus cosecant cotangent. Matches. So look at what that means practically. Only cosh breaks from the trigonometric sign pattern. One row out of six. If you're already fluent in trigonometric derivatives, that third column makes this table nearly automatic. Learn the cosh row — the one exception — and read the rest off what you already know.

Applications

Composites and the Chain Rule

In practice, hyperbolic functions rarely appear with a bare x inside. They're composed with other expressions. So the chain rule supplies an extra factor — the derivative of the inner function. Exactly as it does for trigonometric and exponential composites. Every formula in the table generalizes by replacing x with an inner function u, and multiplying by u prime. First example. Differentiate y equals sinh of three x squared plus one. The outer function is sinh, whose derivative is cosh. The inner function is three x squared plus one, whose derivative is six x. So the answer is six x cosh of three x squared plus one. Second example. Differentiate y equals x squared tanh of five x. This one needs the product rule, with the second factor also needing the chain rule. The derivative of tanh of five x is sech squared of five x, times five. So you get two x tanh five x, plus five x squared sech squared five x. The procedure never changes. Identify outer and inner. Apply the matching derivative from the table. Multiply by the derivative of the inner function.

Two Results Worth Knowing

The Catenary's Slope, and a Surprise

Two more examples, and both give results worth recognizing on sight. Third. A hanging cable follows y equals ten cosh of x over ten. Find its slope, and evaluate at x equals ten. The outer function is cosh, derivative sinh. The inner is x over ten, derivative one over ten. So d y d x is ten, times sinh of x over ten, times one tenth. And look what happens. The ten and the one tenth cancel exactly. The slope is simply sinh of x over ten. At x equals ten, that's sinh of one, which is about one point one seven five. And that cancellation isn't a coincidence of this example. A catenary of the form a cosh of x over a always has slope sinh of x over a. That's a tidy result used directly in cable engineering. Fourth example. Differentiate y equals log of cosh x. The outer function is the natural logarithm, derivative one over u. The inner is cosh x, derivative sinh x. So you get one over cosh x, times sinh x. Which is sinh over cosh. Which is tanh x. The derivative of log cosh x is simply tanh x. An elegant result. And it runs in reverse too — telling you that log cosh x is the antiderivative of tanh x. Results like that make hyperbolic functions valuable for integration as well as differentiation. Which is exactly where we go next.

Key Takeaways

What to Carry Forward

Before the takeaways, one thing to think about. The derivative of cosine x is minus sine x. But the derivative of cosh x is plus sinh x, with no minus at all. Trace that difference back to the definitions. Cosine relates to exponentials with imaginary arguments — e to the i x. Cosh uses real exponentials — e to the x. So ask yourself: why does differentiating e to the minus x, versus e to the minus i x, produce a different sign outcome? And how does that single fact explain the entire pattern of agreements and disagreements across the table? Five takeaways. The derivative of sinh is cosh, and the derivative of cosh is sinh. Both derived directly from the exponential definitions. The cosh row is the only sign break from trigonometry. Everything else matches its circular counterpart exactly. The other four derivatives follow from the quotient and chain rules, and the three carrying a leading minus are the reciprocal-prefixed ones. Composites work exactly as elsewhere. Outer derivative, times inner derivative. And two results worth recognizing on sight: the derivative of log cosh x is tanh x, and a catenary a cosh x over a always has slope sinh x over a. Next up is section three point four — Integrals and Inverse Hyperbolic Functions, where we run every one of these derivatives backwards.

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FAQ

Questions students ask us

About Derivatives of Hyperbolic Functions

What is the derivative of sinh x and cosh x?
The derivative of \sinh x is \cosh x, and the derivative of \cosh x is \sinh x. Start from \sinh x = \frac{e^x - e^{-x}}{2} and use \frac{d}{dx}e^{-x} = -e^{-x}. Differentiating term by term gives \frac{e^x + e^{-x}}{2} = \cosh x. Differentiating \cosh x = \frac{e^x + e^{-x}}{2} the same way gives \frac{e^x - e^{-x}}{2} = \sinh x, and no minus sign appears.
How do you find the derivative of tanh x and the other hyperbolic functions?
Write \tanh x = \frac{\sinh x}{\cosh x} and apply the quotient rule to get \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x}. Since \cosh^2 x - \sinh^2 x = 1, this equals \operatorname{sech}^2 x. The same quotient or chain rule approach gives \frac{d}{dx}\coth x = -\operatorname{csch}^2 x, \frac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x \tanh x, and \frac{d}{dx}\operatorname{csch} x = -\operatorname{csch} x \coth x.
How do hyperbolic derivative signs differ from trig derivative signs?
Two of the six hyperbolic derivatives differ in sign from their trigonometric counterparts. The first pair is \frac{d}{dx}\cosh x = \sinh x versus \frac{d}{dx}\cos x = -\sin x. The second pair is \frac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x \tanh x versus \frac{d}{dx}\sec x = \sec x \tan x. The sinh, tanh, coth, and csch rows match trigonometry in sign. A common mistake is copying the cosine minus sign onto cosh or dropping the minus on sech.
How do you use the chain rule with hyperbolic functions?
Multiply the derivative of the outer hyperbolic function by the derivative of the inner function. For example, \frac{d}{dx}\sinh(3x^2+1) = 6x\cosh(3x^2+1). For a catenary y = 10\cosh(x/10), the slope is 10\sinh(x/10)\cdot\frac{1}{10} = \sinh(x/10), which is about 1.175 at x = 10. Similarly, \frac{d}{dx}\ln(\cosh x) = \frac{\sinh x}{\cosh x} = \tanh x.

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