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

Understand the Problem

The question is asking for the differentiation of the expression 'log(cosh(x)) + (1/2)cosh²(x)' with respect to x. This involves applying the rules of differentiation to each term in the expression.

Answer

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

Steps to Solve

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

To differentiate $log(cosh(x))$, we apply the chain rule. The derivative of $log(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$. Here, $u = cosh(x)$, so:

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

The derivative of $cosh(x)$ is $sinh(x)$. Therefore:

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

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

Next, we differentiate $\frac{1}{2}cosh²(x)$. We again use the chain rule, where $u = cosh(x)$. The derivative of $\frac{1}{2}u^2$ is $u \cdot \frac{du}{dx}$, so we have:

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

As previously mentioned, $\frac{d}{dx}(cosh(x)) = sinh(x)$. Therefore:

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

Combine the derivatives

Now we combine the derivatives of both terms:

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

The derivative of the expression is:

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

More Information

Differentiating logarithmic and hyperbolic functions often involves applying both the chain rule and product rule. The derivative of a logarithm is particularly useful in simplifying hyperbolic expressions.

Tips

Forgetting to apply the chain rule when differentiating composite functions like $log(cosh(x))$.
Confusing the derivatives of hyperbolic functions; ensure you remember that $\frac{d}{dx}(cosh(x)) = sinh(x)$ and $\frac{d}{dx}(sinh(x)) = cosh(x)$.

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