Derivative of cosh(x)

The derivative of the hyperbolic cosine is the hyperbolic sine: $\dfrac{d}{dx}\cosh x = \sinh x$. Note the difference to the ordinary cosine, whose derivative is $-\sin x$.

$$ \frac{d}{dx}\left(\cosh x\right) = \sinh x $$

See the derivative of cosh(x) step by step in the calculator →

📈 Graph of cosh(x) and its derivative

Graph of cosh(x) (solid) and its derivative (dashed) −3−2−10123−5−4−3−2−10123456 f(x) = cosh xf′(x) = sinh x
cosh is the shape of a hanging chain (catenary). Its derivative $\sinh x$ is negative for $x<0$ and positive for $x>0$, so cosh falls, reaches its minimum $1$ at $x=0$ (where $\sinh 0 = 0$) and rises again.

🔍 Why is the derivative of cosh(x) equal to $\sinh x$?

With $\cosh x = \frac{e^x + e^{-x}}{2}$:

$$ \frac{d}{dx}\cosh x = \frac{e^x + (-e^{-x})}{2} = \frac{e^x - e^{-x}}{2} = \sinh x. $$

The minus sign that appears in $(\cos x)'$ has no counterpart here: it comes from $i^2 = -1$ in $\cos x = \frac{e^{ix}+e^{-ix}}{2}$, and there is no $i$ in the hyperbolic case.

🔗 Derivative of cosh(x) with an inner function

If the argument is a function $u(x)$, the chain rule adds the inner derivative:

$$ \frac{d}{dx}\cosh\big(u(x)\big) = u'(x)\sinh\big(u(x)\big) $$

✏️ Worked examples

FunctionDerivativeNote
$\cosh(2x)$$2\sinh(2x)$ steps →
$\cosh(x^2)$$2x\sinh(x^2)$ steps →
$\ln(\cosh x)$$\tanh x$$\frac{\sinh x}{\cosh x}$ by the chain rule. steps →
$e^x\cosh x$$e^x(\cosh x + \sinh x) = e^{2x}$Product rule; use $\cosh x + \sinh x = e^x$. steps →

⚠️ Common mistakes

Writing $-\sinh x$ by analogy with $\cos x$.

Forgetting the inner derivative: $(\cosh 5x)' = 5\sinh 5x$.

❓ Frequently asked questions

Why is the derivative of cosh(x) positive sinh(x)?

Differentiating $e^{-x}$ gives $-e^{-x}$, which turns the sum $e^x + e^{-x}$ into the difference $e^x - e^{-x}$ – that is $2\sinh x$, with no extra sign.

What is the second derivative of cosh(x)?

$\cosh x$ itself.

Where does cosh(x) have its minimum?

At $x = 0$, where the derivative $\sinh x$ is zero; the minimum value is $\cosh 0 = 1$.