Derivative Calculator

Hyperbolic rule

The hyperbolic functions behave like the trigonometric ones, but without the alternating minus signs.

\[\frac{d}{dx}\tanh x = \operatorname{sech}^{2} x\]

What it says

They are built directly from exponentials: \(\sinh x = \frac{e^{x} - e^{-x}}{2}\) and \(\cosh x = \frac{e^{x} + e^{-x}}{2}\). Differentiating those definitions immediately gives \(\left(\sinh x\right)' = \cosh x\) and \(\left(\cosh x\right)' = \sinh x\) — no minus sign anywhere.

The rest follow by the quotient rule: \(\left(\tanh x\right)' = \operatorname{sech}^{2}x\), \(\left(\operatorname{sech} x\right)' = -\operatorname{sech} x\tanh x\), \(\left(\coth x\right)' = -\operatorname{csch}^{2}x\).

The identity in play is \(\cosh^{2}x - \sinh^{2}x = 1\), a minus where the circular functions have a plus. That single sign difference is the whole distinction between the two families.

When it applies

  • \(\sinh\), \(\cosh\), \(\tanh\) and their reciprocals.
  • Catenary curves, and \(\tanh\) as an activation function in neural networks.
  • With the chain rule for a compound argument.

Five worked examples

Every line is the step the calculator would show, in the order it applies them. Each graph is live: hover it to read both curves and see the tangent whose slope is the derivative, drag to pan, scroll to zoom.

Example 1
\[\frac{d}{dx}\left[\sinh\left(x\right)\right]\]

  1. Differentiate the definition term by term.

    \[\frac{d}{dx}\left[\frac{e^{x} - e^{-x}}{2}\right] = \frac{e^{x} + e^{-x}}{2}\]

  2. Which is exactly \(\cosh x\).

    \[= \cosh x\]
Answer
\[\cosh\left(x\right)\]

f′ is \(\cosh\), which sits above \(\sinh\) at every point, so the two curves approach but never meet. To the right both are dominated by \(\tfrac{e^{x}}{2}\) and the gap closes; to the left, \(\sinh\) dives while \(\cosh\) climbs.

No minus sign, unlike the circular case.

Open sinh(x) in the calculator
Example 2
\[\frac{d}{dx}\left[\cosh\left(x\right)\right]\]

  1. Same derivation the other way round.

    \[= \frac{e^{x} - e^{-x}}{2} = \sinh x\]
Answer
\[\sinh\left(x\right)\]

The catenary — the shape of a hanging chain. f′ crosses zero at \(x = 0\), where the chain reaches its lowest point.

The pair differentiates into each other with period two, not four.

Open cosh(x) in the calculator
Example 3
\[\frac{d}{dx}\left[\tanh\left(x\right)\right]\]

  1. Quotient rule on \(\frac{\sinh x}{\cosh x}\).

    \[= \frac{\cosh^{2}x - \sinh^{2}x}{\cosh^{2}x}\]

  2. The identity makes the numerator \(1\).

    \[= \frac{1}{\cosh^{2}x} = \operatorname{sech}^{2}x\]
Answer
\[\operatorname{sech}^{2}\left(x\right)\]

f is a smooth step from \(-1\) to \(1\), steepest at the origin, and f′ is the bump that measures that steepness.

Always between \(0\) and \(1\), and vanishing at both tails — the saturation that makes \(\tanh\) useful as an activation.

Open tanh(x) in the calculator
Example 4
\[\frac{d}{dx}\left[\sinh\left(3 x\right)\right]\]

  1. Chain rule with \(u = 3x\).

    \[= \cosh\left(3x\right)\cdot 3\]
Answer
\[3 \cosh\left(3 x\right)\]

Compressed threefold, so f′ is \(3\cosh 3x\) and towers over f. Both curves leave the window almost at once.

The chain rule behaves exactly as it does for the circular functions.

Open sinh(3x) in the calculator
Example 5
\[\frac{d}{dx}\left[x \cosh\left(x\right)\right]\]

  1. Product rule.

    \[= 1 \cdot \cosh x + x\sinh x\]
Answer
\[\cosh\left(x\right) + x \sinh\left(x\right)\]

f rises through the origin and f′ stays positive everywhere, so the curve never turns.

The shape of a hanging chain weighted along its length.

Open x cosh(x) in the calculator

Watch out for

  • \(\left(\cosh x\right)' = +\sinh x\). The minus sign of the circular case is absent here.
  • \(\operatorname{sech}\) and \(\operatorname{csch}\) do pick up minus signs — the pattern is not "no minus signs anywhere".
  • \(\cosh^{2}x - \sinh^{2}x = 1\), not \(+\). Using the circular identity gives the wrong numerator.
Exponential ruleInverse hyperbolic ruleTrigonometric rule