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

Steps to Solve

1. Identify the Function for Differentiation

We need to differentiate the function:
$$ f(x) = \log(\cosh(x)) + \frac{1}{2\cosh^2(x)} $$

2. Differentiate the First Term

To differentiate the first term $ \log(\cosh(x)) $, we will use the chain rule:

• The derivative of $ \log(u) $ is $\frac{1}{u} \cdot \frac{du}{dx}$, where $ u = \cosh(x) $.
• The derivative of $ u = \cosh(x) $ is $\sinh(x) $. Thus, we have: $$ \frac{d}{dx} \log(\cosh(x)) = \frac{1}{\cosh(x)} \cdot \sinh(x) = \frac{\sinh(x)}{\cosh(x)} $$

3. Differentiate the Second Term

Next, we differentiate the second term $ \frac{1}{2\cosh^2(x)} $: This can be viewed as a composition of functions that requires the chain rule: Let $ v = \cosh(x)$, so our term becomes $ \frac{1}{2v^2} $.

• The derivative of $ \frac{1}{2v^2} $ using the power rule is $ -\frac{1}{v^3} \cdot \frac{dv}{dx} $, where $ \frac{dv}{dx} = \sinh(x) $. Thus, we get: $$ \frac{d}{dx} \left( \frac{1}{2\cosh^2(x)} \right) = -\frac{1}{2 \cosh^3(x)} \cdot \sinh(x) $$

4. Combine the Results

Now we combine the results from both derivatives to find the overall derivative: $$ f'(x) = \frac{\sinh(x)}{\cosh(x)} - \frac{1}{2 \cosh^3(x)} \sinh(x) $$

5. Factor Out Common Terms

Both terms have $\sinh(x)$ in common: $$ f'(x) = \sinh(x) \left( \frac{1}{\cosh(x)} - \frac{1}{2\cosh^3(x)} \right) $$

6. Final Simplification

To reach the final expression, we can combine the terms inside the parentheses: $$ f'(x) = \sinh(x) \left( \frac{2}{2\cosh(x)} - \frac{1}{2\cosh^3(x)} \right) = \sinh(x) \left( \frac{2\cosh^2(x)-1}{2\cosh^3(x)} \right) $$

Thus, we conclude that: $$ f'(x) = \frac{\sinh(x)(2\cosh^2(x)-1)}{2\cosh^3(x)} $$