## Proof

The hyperbolic cosine function is defined as:

$$\mathrm{cosh}(x)=\frac{{e}^{x}+{e}^{-x}}{2}$$To find the derivative, we differentiate using the sum rule:

$$\frac{\text{d}}{\text{d}x}\mathrm{cosh}(x)=\frac{\text{d}}{\text{d}x}\left(\frac{{e}^{x}+{e}^{-x}}{2}\right)$$Differentiating each term:

$$\frac{1}{2}(\frac{\text{d}}{\text{d}x}{e}^{x}+\frac{\text{d}}{\text{d}x}{e}^{-x})$$ $$=\frac{1}{2}({e}^{x}-{e}^{-x})$$This simplifies to:

$$\mathrm{sinh}(x)$$Thus, the derivative of $\mathrm{cosh}(x)$ is:

$$\frac{\text{d}}{\text{d}x}\mathrm{cosh}(x)=\overline{)\mathrm{sinh}(x)}$$## Explanation

The hyperbolic cosine function, $\mathrm{cosh}(x)$, is defined as $\frac{{e}^{x}+{e}^{-x}}{2}$. This formula combines the exponential functions ${e}^{x}$ and ${e}^{-x}$.

To differentiate $\mathrm{cosh}(x)$, we use basic differentiation rules. The function can be broken down into $\frac{1}{2}({e}^{x}+{e}^{-x})$. Here, $\frac{1}{2}$ is a constant factor that we can factor out during differentiation.

We then apply the differentiation rule to each part of the expression. The derivative of ${e}^{x}$ is simply ${e}^{x}$, and the derivative of ${e}^{-x}$ is $-{e}^{-x}$ due to the chain rule.

Putting these results together, the derivative becomes $\frac{1}{2}({e}^{x}-{e}^{-x})$. This expression is the definition of $\mathrm{sinh}(x)$, the hyperbolic sine function. Therefore, the derivative of cosh(x) is sinh(x).

*Q.E.D.*