1. Derivatives of hyperbolic functions
Recall from Chapter 1 that hyperbolic functions are built directly from
e^x:
Because these are just combinations of exponentials, and we already know
d/dx(eˣ) = eˣ and d/dx(e⁻ˣ) = −e⁻ˣ,
differentiating sinh and cosh
is almost embarrassingly easy — and this is exactly why they swap into each other
instead of picking up a minus sign like sin and cos do.
Derivation
→ Start with y = sinh x = ½(eˣ − e⁻ˣ)
→ Differentiate term by term: dy/dx = ½(eˣ − (−e⁻ˣ)) = ½(eˣ + e⁻ˣ)
→ That's exactly the definition of cosh x.
→ d/dx(sinh x) = cosh x
→ Do the same for cosh x = ½(eˣ + e⁻ˣ): differentiating gives ½(eˣ − e⁻ˣ)
→ d/dx(cosh x) = sinh x
Now for tanh x, we use the quotient rule on
sinh x / cosh x:
Derivation
→ Quotient rule: d/dx(sinh x / cosh x) = [cosh x · cosh x − sinh x · sinh x] / cosh²x
→ Numerator: cosh²x − sinh²x = 1 (the hyperbolic Pythagorean identity from Ch.1)
→ So we're left with 1/cosh²x = sech²x
→ d/dx(tanh x) = sech²x
The other three — coth x, sech x, cosech x —
all come from writing them as reciprocals or quotients of sinh/cosh and applying
the quotient rule the same way. You won't usually be asked to reproduce these
derivations in an exam, but you are expected to know the results, so here's
the full table:
| f(x) | f′(x) | compare to trig |
|---|
| sinh x | cosh x | same pattern as sin → cos |
| cosh x | sinh x | ⚠ trig has cos → −sin, this one has no minus |
| tanh x | sech²x | same pattern as tan → sec² |
| coth x | −cosech²x | same pattern as cot → −cosec² |
| sech x | −sech x tanh x | ⚠ trig has sec → +sec tan, this one has an extra minus |
| cosech x | −cosech x coth x | same pattern as cosec → −cosec cot |
⚡ The one thing to actually memorise here
Four of the six hyperbolic derivatives are identical in sign to their trig
equivalents. Only cosh and sech break the pattern — cosh loses its
minus sign, and sech gains one it "shouldn't" have. If you can remember just
those two exceptions, you can reconstruct the whole table from your trig knowledge.
Combining with chain rule, product rule, quotient rule
Hyperbolic functions behave completely normally inside the usual rules —
there's no special trick, just careful bookkeeping.
Worked example — chain rule
Differentiate y = cosh(3x² + 1).
→ Outer function cosh(u) differentiates to sinh(u); inner function u = 3x²+1 differentiates to 6x.
→ dy/dx = 6x sinh(3x² + 1)
Worked example — product rule
Differentiate y = x² tanh x.
→ Product rule: dy/dx = (2x)(tanh x) + (x²)(sech²x)
→ dy/dx = 2x tanh x + x² sech²x
Worked example — a useful "hidden" chain rule
Differentiate y = ln(cosh x).
→ Chain rule: dy/dx = (1/cosh x) × sinh x
→ dy/dx = tanh x
→ Worth remembering — this exact combination shows up again when we integrate tanh x in Chapter 4 (reverse direction!).
Practice 1
Differentiate y = sinh(5x) − 3cosh(2x).
Practice 2
Differentiate y = tanh³x (that is, (tanh x)³).
2. Derivatives of inverse functions
Here's the one really important technique in this whole chapter, and it's a
single reusable idea: to differentiate any inverse function, don't try to
differentiate the inverse directly — flip it round, differentiate implicitly,
then substitute back. This works identically whether you're inverting a
trig function or a hyperbolic one.
Derivation — d/dx(arsinh x)
→ Let y = arsinh x. By definition of an inverse function, this means sinh y = x.
→ Differentiate both sides with respect to x (implicit differentiation — y is a function of x):
cosh y · dy/dx = 1
→ Rearrange: dy/dx = 1/cosh y
→ Problem: the answer is in terms of y, but we want it in terms of x. Use the identity cosh²y − sinh²y = 1, so cosh y = √(1 + sinh²y).
→ But sinh y = x (from the very first line!) so cosh y = √(1 + x²)
→ d/dx(arsinh x) = 1/√(1 + x²)
💡 Why this always works
Every inverse function derivation follows this exact four-step recipe:
(1) flip to the direct form, (2) differentiate implicitly,
(3) you'll have dy/dx in terms of a function of y,
(4) use a Pythagorean-type identity to swap that for a function of x,
because you know the original variable (sin y, cosh y, etc.) equals x by definition. Learn the recipe, not just the results.
Applying exactly this method to every trig and hyperbolic inverse gives the
full set below. You've already met arcsin, arccos, arctan
from Core/Further Pure — they're included here so you can see how naturally the
hyperbolic ones sit alongside them.
| f(x) | f′(x) |
|---|
| arcsin x | 1/√(1 − x²) |
| arccos x | −1/√(1 − x²) |
| arctan x | 1/(1 + x²) |
| arsinh x | 1/√(1 + x²) |
| arcosh x | 1/√(x² − 1) |
| artanh x | 1/(1 − x²) |
Notice the pairing: arsinh and arcsin
have almost the same derivative — the only difference is a +
instead of a − under the square root. Same story for
arcosh vs (the less common) inverse secant-type form, and for
artanh vs arctan — again just a sign
swap on the x² term. That's Osborn's rule showing up again, quietly, in the denominators.
Derivation — d/dx(arcosh x)
→ Let y = arcosh x, so cosh y = x.
→ Differentiate implicitly: sinh y · dy/dx = 1
→ dy/dx = 1/sinh y, and since cosh²y − sinh²y = 1, we get sinh y = √(cosh²y − 1) = √(x² − 1)
→ d/dx(arcosh x) = 1/√(x² − 1)
Derivation — d/dx(artanh x)
→ Let y = artanh x, so tanh y = x.
→ Differentiate implicitly: sech²y · dy/dx = 1
→ dy/dx = 1/sech²y, and since 1 − tanh²y = sech²y, this is 1/(1 − tanh²y) = 1/(1 − x²)
→ d/dx(artanh x) = 1/(1 − x²)
→ Sanity check using the log form artanh x = ½ln((1+x)/(1−x)): differentiating gives ½[1/(1+x) + 1/(1−x)] = ½ · 2/(1−x²) = 1/(1−x²). ✓ Same answer, two routes.
Just like before, these plug straight into the chain rule and product rule
for anything more complicated:
Worked example — chain rule
Differentiate y = arsinh(2x).
→ Chain rule: outer derivative is 1/√(1 + u²) with u = 2x; inner derivative is 2.
→ dy/dx = 2/√(1 + 4x²)
Worked example — product rule
Differentiate y = x·arcosh x.
→ Product rule: dy/dx = (1)(arcosh x) + (x)(1/√(x²−1))
→ dy/dx = arcosh x + x/√(x² − 1)
→ This exact combination is the building block for integrating arcosh x by parts in Chapter 4 — keep it in mind.
Practice 3
Find dy/dx when y = arcosh(3x).
Practice 4
Differentiate y = artanh(x/2).
Practice 5 — derive it yourself
Using the method shown above, show that d/dx(coth x) = −cosech²x.
Exam Tips
🚨 Trap #1 — the cosh/sech sign flip
This is the single most common slip in this chapter. Students who know their
trig derivatives well instinctively write d/dx(cosh x) = −sinh x
or d/dx(sech x) = sech x tanh x because that's the trig
pattern. Both are wrong. Say it out loud until it sticks: "cosh loses its minus,
sech gains one."
🚨 Trap #2 — forgetting the chain rule multiplier
When differentiating something like arsinh(5x) or
cosh(x² − 1), it's easy to write down the "outer"
derivative and forget to multiply by the derivative of the inner function.
Examiners deliberately choose inner functions that aren't just x
to test exactly this.
✅ Good habit — check your identity direction
When substituting back from a function of y to a function
of x, make sure you're using the identity the right way round.
For arcosh you need sinh y = √(cosh²y − 1)
(positive root, since arcosh x is defined for x ≥ 1
where sinh y ≥ 0). Getting the sign of the square root wrong
is an easy way to lose a mark even when your method is otherwise perfect.
📝 What examiners are looking for
Full marks on a "differentiate y = ..." question almost always require: correctly
identifying which rule(s) apply (chain / product / quotient), quoting the correct
standard derivative from the tables above, and simplifying the final answer where
possible (e.g. combining fractions, factorising). Marks are often given for method
even if the final simplification isn't perfect — so show your working line by line,
the way the worked examples above do.
🔮 Looking ahead
Every derivative in this chapter reappears backwards in Chapter 4 (Integration) —
each "d/dx(...)" result becomes a "∫...dx" result read right-to-left. If this
chapter feels solid, integration will feel like recognition rather than new learning.