Derivatives of Hyperbolic Functions for MATH 139
Exam Relevance for MATH 139
Coverage varies by instructor. If covered, know sinh and cosh derivatives and their similarity to trig.
Derivatives of Hyperbolic Functions
Hyperbolic functions are relatives of the regular trig functions, but they're defined using exponentials. The good news: their derivatives are simpler and more symmetrical than regular trig!
What Are Hyperbolic Functions?
The hyperbolic functions are defined in terms of $e^x$:
$$\sinh x = \frac{e^x - e^{-x}}{2} \qquad \cosh x = \frac{e^x + e^{-x}}{2}$$
$$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$
🔑 Pronunciation:
- $\sinh$ = "sinch" or "shine"
- $\cosh$ = "cosh"
- $\tanh$ = "tanch" or "than"
The Beautiful Symmetry
Unlike regular trig, hyperbolic derivatives are almost perfectly symmetrical:
$$\frac{d}{dx}[\sinh x] = \cosh x$$
$$\frac{d}{dx}[\cosh x] = \sinh x$$
Notice: No negative sign! Compare to regular trig where $\frac{d}{dx}[\cos x] = -\sin x$.
All Six Derivatives
| Function | Derivative |
|---|---|
| $\sinh x$ | $\cosh x$ |
| $\cosh x$ | $\sinh x$ |
| $\tanh x$ | $\text{sech}^2 x$ |
| $\coth x$ | $-\text{csch}^2 x$ |
| $\text{sech } x$ | $-\text{sech } x \tanh x$ |
| $\text{csch } x$ | $-\text{csch } x \coth x$ |
Pattern: Only $\sinh$ and $\cosh$ have positive derivatives. All the others (including the "co-" functions) are negative.
Comparison: Trig vs Hyperbolic
| Regular Trig | Hyperbolic |
|---|---|
| $\frac{d}{dx}[\sin x] = \cos x$ | $\frac{d}{dx}[\sinh x] = \cosh x$ |
| $\frac{d}{dx}[\cos x] = -\sin x$ | $\frac{d}{dx}[\cosh x] = \sinh x$ ✨ |
| $\frac{d}{dx}[\tan x] = \sec^2 x$ | $\frac{d}{dx}[\tanh x] = \text{sech}^2 x$ |
The big difference: $\cosh$ derivative has no negative sign!
Problem: Find $\frac{d}{dx}[3\sinh x + 2\cosh x]$
$$\frac{d}{dx}[3\sinh x + 2\cosh x] = 3\cosh x + 2\sinh x$$
$$= \boxed{3\cosh x + 2\sinh x}$$
Problem: Prove that $\frac{d}{dx}[\sinh x] = \cosh x$ using the definition.
Step 1: Write out the definition: $$\sinh x = \frac{e^x - e^{-x}}{2}$$
Step 2: Differentiate: $$\frac{d}{dx}[\sinh x] = \frac{1}{2}\frac{d}{dx}[e^x - e^{-x}]$$ $$= \frac{1}{2}[e^x - (-1)e^{-x}]$$ $$= \frac{1}{2}[e^x + e^{-x}]$$
Step 3: Recognize the result: $$= \frac{e^x + e^{-x}}{2} = \boxed{\cosh x}$$
Problem: Find $\frac{d}{dx}[x^2 \sinh x]$
Step 1: Apply the product rule: $$\frac{d}{dx}[x^2 \sinh x] = \frac{d}{dx}[x^2] \cdot \sinh x + x^2 \cdot \frac{d}{dx}[\sinh x]$$
Step 2: Calculate: $$= 2x \cdot \sinh x + x^2 \cdot \cosh x$$
$$= \boxed{2x\sinh x + x^2\cosh x}$$
Problem: Find $\frac{d}{dx}[\cosh(3x)]$
Step 1: Apply the chain rule: $$\frac{d}{dx}[\cosh(3x)] = \sinh(3x) \cdot \frac{d}{dx}[3x]$$
Step 2: Simplify: $$= \sinh(3x) \cdot 3 = \boxed{3\sinh(3x)}$$
Problem: Prove that $\frac{d}{dx}[\tanh x] = \text{sech}^2 x$
Step 1: Write $\tanh x = \frac{\sinh x}{\cosh x}$ and apply quotient rule:
$$\frac{d}{dx}[\tanh x] = \frac{\cosh x \cdot \cosh x - \sinh x \cdot \sinh x}{\cosh^2 x}$$
Step 2: Simplify the numerator: $$= \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x}$$
Step 3: Use the hyperbolic identity $\cosh^2 x - \sinh^2 x = 1$: $$= \frac{1}{\cosh^2 x} = \boxed{\text{sech}^2 x}$$
Problem: Find where $f(x) = \sinh x - 2x$ has horizontal tangent lines.
Step 1: Set $f'(x) = 0$: $$f'(x) = \cosh x - 2 = 0$$ $$\cosh x = 2$$
Step 2: Solve for $x$.
Recall $\cosh x = \frac{e^x + e^{-x}}{2}$, so: $$\frac{e^x + e^{-x}}{2} = 2$$ $$e^x + e^{-x} = 4$$
Let $u = e^x$: $$u + \frac{1}{u} = 4$$ $$u^2 - 4u + 1 = 0$$ $$u = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$$
Step 3: Since $u = e^x > 0$, both solutions are valid: $$x = \ln(2 + \sqrt{3}) \quad \text{or} \quad x = \ln(2 - \sqrt{3})$$
Note: $\ln(2 - \sqrt{3}) = -\ln(2 + \sqrt{3})$ (these are negatives of each other).
$$\boxed{x = \pm\ln(2 + \sqrt{3})}$$
Key Hyperbolic Identities
These are analogous to trig identities:
$$\cosh^2 x - \sinh^2 x = 1$$
$$1 - \tanh^2 x = \text{sech}^2 x$$
$$\coth^2 x - 1 = \text{csch}^2 x$$
Notice the sign difference from trig: it's $\cosh^2 - \sinh^2$ (not plus)!
Common Mistakes and Misunderstandings
❌ Mistake: Adding a negative sign to $\frac{d}{dx}[\cosh x]$
Wrong: $\frac{d}{dx}[\cosh x] = -\sinh x$
Why it's wrong: Unlike $\cos x$, the derivative of $\cosh x$ has NO negative sign.
Correct: $\frac{d}{dx}[\cosh x] = \sinh x$
❌ Mistake: Confusing hyperbolic and regular trig
Wrong: Thinking $\sinh^2 x + \cosh^2 x = 1$
Why it's wrong: The hyperbolic identity has a MINUS sign: $\cosh^2 x - \sinh^2 x = 1$
Correct: Remember: hyperbolic functions are different from regular trig!
❌ Mistake: Forgetting chain rule
Wrong: $\frac{d}{dx}[\sinh(5x)] = \cosh(5x)$
Why it's wrong: You must multiply by the derivative of the inside.
Correct: $\frac{d}{dx}[\sinh(5x)] = 5\cosh(5x)$
Quick Memory Aid
The "nice" pair: sinh and cosh
- $\sinh' = \cosh$ (positive)
- $\cosh' = \sinh$ (positive, no negative!)
Everything else is negative:
- $\tanh' = \text{sech}^2$ (but sech derivatives are negative)
- All "co-" and "sech/csch" derivatives have minus signs
Derivative of Sinh
The derivative of hyperbolic sine is hyperbolic cosine. One of the two fundamental hyperbolic derivatives.
Variables:
- $x$:
- any real number
Derivative of Cosh
The derivative of hyperbolic cosine is hyperbolic sine. NO negative sign (unlike regular cosine)!
Variables:
- $x$:
- any real number
Derivative of Tanh
The derivative of hyperbolic tangent is hyperbolic secant squared.
Variables:
- $x$:
- any real number
Derivative of Coth
The derivative of hyperbolic cotangent is NEGATIVE hyperbolic cosecant squared.
Variables:
- $x$:
- any real number except 0
Derivative of Sech
The derivative of hyperbolic secant. Note the negative sign.
Variables:
- $x$:
- any real number
Derivative of Csch
The derivative of hyperbolic cosecant. Note the negative sign.
Variables:
- $x$:
- any real number except 0
Fundamental Hyperbolic Identity
The hyperbolic Pythagorean identity. Note it's MINUS (not plus like regular trig).
Variables:
- $x$:
- any real number
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