Derivatives of Hyperbolic Functions for Master Mathematics
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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