Derivative of sinh(x)

The derivative of the hyperbolic sine is the hyperbolic cosine: $\dfrac{d}{dx}\sinh x = \cosh x$. Unlike $\sin x$, there are no sign changes when you keep differentiating.

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

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

📈 Graph of sinh(x) and its derivative

Graph of sinh(x) (solid) and its derivative (dashed) −3−2−10123−5−4−3−2−10123456 f(x) = sinh xf′(x) = cosh x
Since $\cosh x \ge 1$ for every $x$, sinh is strictly increasing and its slope is never smaller than $1$ – the minimum slope $1$ occurs at the origin. For large $|x|$ both curves approach $\pm\frac{e^{|x|}}{2}$.

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

From the definition $\sinh x = \frac{e^x - e^{-x}}{2}$ and $(e^{-x})' = -e^{-x}$:

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

Differentiating again gives $(\cosh x)' = \sinh x$, so the derivatives alternate between sinh and cosh: the $n$-th derivative is $\sinh x$ for even $n$ and $\cosh x$ for odd $n$.

🔗 Derivative of sinh(x) with an inner function

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

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

✏️ Worked examples

FunctionDerivativeNote
$\sinh(3x)$$3\cosh(3x)$Inner derivative $3$. steps →
$\sinh(x^2)$$2x\cosh(x^2)$ steps →
$\sinh^2 x$$2\sinh x\cosh x = \sinh(2x)$Power of a function, then the double-angle identity. steps →
$x\sinh x$$\sinh x + x\cosh x$Product rule. steps →

⚠️ Common mistakes

Adding a minus sign out of habit from $(\cos x)' = -\sin x$: hyperbolic derivatives of sinh and cosh have none.

Writing $(\sinh x)' = \sinh x$ – that is true for $e^x$, not for sinh.

❓ Frequently asked questions

What is the second derivative of sinh(x)?

$\sinh x$ again: $(\sinh x)'' = (\cosh x)' = \sinh x$.

Is the derivative of sinh(x) equal to cosh(x)?

Yes, $\frac{d}{dx}\sinh x = \cosh x$, with a plus sign.

What is the derivative of sinh(2x)?

$2\cosh(2x)$.