Derivative Calculator

Derivatives of Hyperbolic Functions (sinh, cosh, tanh)

Hyperbolic functions are the exponential counterparts of trigonometric functions. Their derivatives follow simple, elegant patterns and extend naturally via the chain rule to composite inputs.

📜 Core Formulas

$$ \frac{d}{dx}\big(\sinh x\big) = \cosh x $$ $$ \frac{d}{dx}\big(\cosh x\big) = \sinh x $$ $$ \frac{d}{dx}\big(\tanh x\big) = \operatorname{sech}^2 x = \frac{1}{\cosh^2 x} $$

💡 Useful identities: $\cosh^2 x - \sinh^2 x = 1$, $\tanh x = \dfrac{\sinh x}{\cosh x}$, $\operatorname{sech} x = \dfrac{1}{\cosh x}$.
🧠 Analogy: As with sin/cos, repeated differentiation produces cycles, but note there’s no sign flip between $\cosh$ and $\sinh$.

🪜 Chain-Rule Application

  1. Identify the outer function: $\sinh(\,\cdot\,)$, $\cosh(\,\cdot\,)$, or $\tanh(\,\cdot\,)$.
  2. Differentiate the outer: $\dfrac{d}{du}\sinh u=\cosh u$, $\dfrac{d}{du}\cosh u=\sinh u$, $\dfrac{d}{du}\tanh u=\operatorname{sech}^2 u$.
  3. Multiply by the inner derivative: Multiply by $u'(x)$.
$$ \frac{d}{dx}\big(\sinh(u(x))\big) = \cosh(u(x))\,u'(x) $$ $$ \frac{d}{dx}\big(\cosh(u(x))\big) = \sinh(u(x))\,u'(x) $$ $$ \frac{d}{dx}\big(\tanh(u(x))\big) = \operatorname{sech}^2(u(x))\,u'(x) $$

✏️ Examples

Example 1: Basic derivatives

  • $\dfrac{d}{dx}\big(\sinh x\big)=\cosh x$
  • $\dfrac{d}{dx}\big(\cosh x\big)=\sinh x$
  • $\dfrac{d}{dx}\big(\tanh x\big)=\dfrac{1}{\cosh^2 x}$

Example 2: Chain rule with a linear argument

$f(x)=\sinh(3x-2)$

$$ f'(x)=\cosh(3x-2)\cdot 3 = 3\,\cosh(3x-2). $$

Example 3: Chain rule with a square

$g(x)=\cosh(x^2)$

$$ g'(x)=\sinh(x^2)\cdot 2x = 2x\,\sinh(x^2). $$

Example 4: Quotient form of $\tanh'$

$h(x)=\tanh(u(x))$

$$ h'(x)=\operatorname{sech}^2(u(x))\,u'(x)=\frac{u'(x)}{\cosh^2(u(x))}. $$

Example 5: Product rule combo

$p(x)=x\cdot \sinh x$

$$ p'(x)=\sinh x + x\cosh x. $$