Derivative of sinh(x) - Proof and Explanation

Proof

The hyperbolic sine function is defined as:

\sinh (x) = \frac{e^{x} - e^{- x}}{2}

To find the derivative, we use the derivative of exponential functions. The derivative of $e^{x}$ is $e^{x}$ , and the derivative of $e^{- x}$ is $- e^{- x}$ .

Now, differentiating $\sinh (x)$ :

\frac{d}{d x} (\frac{e^{x} - e^{- x}}{2}) = \frac{1}{2} (e^{x} - (- e^{- x}))

This simplifies to:

\frac{1}{2} (e^{x} + e^{- x}) = \cosh (x)

Thus, the derivative of $\sinh (x)$ is:

\frac{d}{d x} \sinh (x) = \cosh (x)

Explanation

The hyperbolic sine function, $\sinh (x)$ , is similar to the sine function but based on exponential functions. It is defined as:

\sinh (x) = \frac{e^{x} - e^{- x}}{2}

This expression represents the difference between the exponential growth $e^{x}$ and the exponential decay $e^{- x}$ , divided by two.

To find the derivative, we differentiate each part of the function. The derivative of $e^{x}$ with respect to $x$ is $e^{x}$ , and the derivative of $e^{- x}$ with respect to $x$ is $- e^{- x}$ . This is because of the chain rule, where the derivative of $- x$ is $- 1$ .

Substituting these derivatives into the formula for $\sinh (x)$ :

\frac{d}{d x} (\frac{e^{x} - e^{- x}}{2}) = \frac{1}{2} (e^{x} - (- e^{- x}))

This simplifies to:

\frac{1}{2} (e^{x} + e^{- x})

This expression is exactly the definition of $\cosh (x)$ , the hyperbolic cosine function:

\frac{e^{x} + e^{- x}}{2} = \cosh (x)

Q.E.D.