📐 Quick Formula Reference
\(\sinh x = \dfrac{e^x - e^{-x}}{2}\)
\(\cosh x = \dfrac{e^x + e^{-x}}{2}\)
\(\tanh x = \dfrac{\sinh x}{\cosh x}\)
\(\cosh^2 x - \sinh^2 x = 1\)
\(\frac{d}{dx}\sinh x = \cosh x\)
\(\frac{d}{dx}\cosh x = \sinh x\)
\(\frac{d}{dx}\tanh x = \text{sech}^2 x\)
\(\sinh^{-1}x = \ln(x+\sqrt{x^2+1})\)
\(\cosh^{-1}x = \ln(x+\sqrt{x^2-1}),\ x\ge1\)
\(\tanh^{-1}x = \frac{1}{2}\ln\!\frac{1+x}{1-x},\ |x|<1\)
Chapter 1 — Definitions & Basic Identities
Memory Key · "Even/Odd Split"
cosh = even part of \(e^x\) · sinh = odd part of \(e^x\)
Think: Cosh is Common (symmetric), sinh is skewed (anti-symmetric).
Q 01
Which of the following correctly expresses \(\sinh x\) using exponentials?
⚠️ Trick: \(\cosh\) and \(\sinh\) look very similar — the sign in the numerator is the only difference.
Q 02
Which fundamental identity is the hyperbolic analogue of \(\cos^2\theta + \sin^2\theta = 1\)?
⚠️ Trick: Unlike trig, the sign flips — it's a DIFFERENCE, not a sum, and \(\cosh\) comes first.
Q 03
Which of the following statements about \(\sinh x\) and \(\cosh x\) is correct?
Q 04
Simplify: \(\cosh^2 x + \sinh^2 x\).
Hint: use the Pythagorean identity and the double-angle formula.
Hint: use the Pythagorean identity and the double-angle formula.
⚠️ Trick: Students often answer "1" — that's \(\cosh^2 x - \sinh^2 x\). The sum is different!
Q 05
What is \(\sinh(x+y)\)?
⚠️ Trick: Looks like the trig formula, but all signs stay positive — no minus sign in hyperbolic addition!
Chapter 2 — Derivatives
Memory Key · "No Negatives"
d/dx sinh = cosh · d/dx cosh = sinh · d/dx tanh = sech²
Unlike trig (\(\frac{d}{dx}\cos = -\sin\)), hyperbolic derivatives never introduce a minus sign in the basic three.
Q 06
What is \(\dfrac{d}{dx}\cosh(3x^2)\)?
Q 07
Differentiate \(f(x) = \tanh^2(x)\). Which answer is correct?
⚠️ Trick: Apply chain rule — \(\tanh^2\) means \([\tanh]^2\), so differentiate the outer power first, then multiply by the derivative of \(\tanh\).
Q 08
To prove \(\dfrac{d}{dx}\tanh x = \text{sech}^2 x\) using the quotient rule, the key step is to evaluate:
\[\frac{d}{dx}\left(\frac{\sinh x}{\cosh x}\right) = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x}\]
What identity makes the numerator simplify to 1?
Q 09
Find the derivative of \(y = x\sinh x - \cosh x\).
⚠️ Trick: Use product rule on \(x\sinh x\), then differentiate \(-\cosh x\). Watch signs carefully.
Q 10
What is \(\dfrac{d}{dx}\left[\ln(\cosh x)\right]\)?
Chapter 3 — Integrals
Memory Key · "Flip the Derivative"
∫ sinh x dx = cosh x + C · ∫ cosh x dx = sinh x + C · ∫ sech²x dx = tanh x + C
Just reverse the derivative table — and no minus signs appear in any of these.
Q 11
Evaluate \(\displaystyle\int \sinh(5x)\,dx\).
Q 12
Evaluate \(\displaystyle\int \tanh x\,dx\).
(Write \(\tanh x = \dfrac{\sinh x}{\cosh x}\) and use substitution.)
⚠️ Trick: Let \(u = \cosh x\), so \(du = \sinh x\,dx\). The integral becomes \(\int \frac{du}{u}\).
Q 13
Evaluate \(\displaystyle\int_0^{\ln 2} \cosh x\,dx\).
⚠️ Trick: \(\cosh(\ln 2) = \dfrac{2 + \tfrac{1}{2}}{2} = \dfrac{5}{4}\) and \(\sinh(0) = 0\).
Q 14
Evaluate \(\displaystyle\int \sinh^2 x\,dx\).
Use the identity \(\sinh^2 x = \dfrac{\cosh(2x)-1}{2}\).
Use the identity \(\sinh^2 x = \dfrac{\cosh(2x)-1}{2}\).
Q 15
Evaluate \(\displaystyle\int \dfrac{1}{\sqrt{x^2+1}}\,dx\).
This is a standard result related to inverse hyperbolic functions.
This is a standard result related to inverse hyperbolic functions.
⚠️ Trick: \(\dfrac{d}{dx}\sinh^{-1}x = \dfrac{1}{\sqrt{x^2+1}}\), so the answer is just the inverse!
Chapter 4 — Inverse Hyperbolic Functions
Memory Key · "Logarithm In Disguise"
Every inverse hyperbolic function is expressible as a ln — memorize the three:
sinh⁻¹x = ln(x+√(x²+1)) · cosh⁻¹x = ln(x+√(x²−1)) · tanh⁻¹x = ½ln((1+x)/(1−x))
Q 16
What is the domain of \(\cosh^{-1} x\)?
⚠️ Trick: \(\cosh x \ge 1\) for all real \(x\), so its inverse is only defined for inputs \(\ge 1\).
Q 17
What is \(\dfrac{d}{dx}\tanh^{-1}(x)\)?
⚠️ Trick: Domain is \(|x|<1\). Students confuse this with \(\coth^{-1}\), which applies for \(|x|>1\) but has the same formula!
Q 18
Which of the following equals \(\tanh^{-1}\!\left(\dfrac{1}{2}\right)\)?
Chapter 5 — Applications & Proofs
Memory Key · "Catenary = cosh"
A hanging chain (catenary) has shape \(y = a\cosh(x/a)\). The arc length formula uses \(\sqrt{1 + \sinh^2 x} = \cosh x\) — the Pythagorean identity in action!
sech²x + tanh²x = 1 is the "other" Pythagorean identity.
Q 19
To prove \(\tanh^2 x + \text{sech}^2 x = 1\), which starting identity should you divide by?
⚠️ Trick: This is the "other" Pythagorean identity — derived by dividing the main identity by \(\cosh^2 x\).
Q 20
The arc length of \(y = \cosh x\) from \(x=0\) to \(x=a\) is given by \(\displaystyle\int_0^a \sqrt{1 + (y')^2}\,dx\). Which expression correctly simplifies the integrand \(\sqrt{1+\sinh^2 x}\)?
⚠️ Trick: Apply the Pythagorean identity \(\cosh^2 x - \sinh^2 x = 1 \Rightarrow 1 + \sinh^2 x = \cosh^2 x\).
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Quiz Complete!
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