Differentiate log(cosh(x)) + 1/2 cosh²(x) with respect to x.

Steps to Solve

1. Differentiate the first term: log(cosh(x))

To find the derivative of $log(cosh(x))$, we use the chain rule. The derivative of $log(u)$ is $\frac{1}{u}$ multiplied by the derivative of $u$. Here, $u = cosh(x)$.

$$ \frac{d}{dx} [log(cosh(x))] = \frac{1}{cosh(x)} \cdot \frac{d}{dx}[cosh(x)] $$

Since the derivative of $cosh(x)$ is $sinh(x)$, we have:

$$ \frac{d}{dx} [log(cosh(x))] = \frac{sinh(x)}{cosh(x)} $$

This simplifies to:

$$ \tanh(x) $$

2. Differentiate the second term: 1/2 cosh²(x)

For this term, we use the power rule and the chain rule. The derivative of $\frac{1}{2} u^2$ is $u \cdot \frac{du}{dx}$. Here, $u = cosh(x)$.

Thus:

$$ \frac{d}{dx} \left[\frac{1}{2} cosh²(x)\right] = cosh(x) \cdot \frac{d}{dx} [cosh(x)] $$

Using the derivative of $cosh(x)$, we find:

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

So the derivative of this term simplifies to:

$$ \frac{1}{2} \cdot 2 \cdot cosh(x) \cdot sinh(x) = cosh(x) \cdot sinh(x) $$

3. Combine the derivatives

Now, we can combine the results from both derivatives:

$$ \frac{d}{dx} [log(cosh(x)) + \frac{1}{2} cosh²(x)] = \tanh(x) + cosh(x) \cdot sinh(x) $$

4. Final expression

The final result for the derivative of the function is:

$$ \tanh(x) + cosh(x) \cdot sinh(x) $$