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

Steps to Solve

1. Differentiate the first term

We start with the function ( \log(\cosh(x)) ). The derivative of ( \log(u) ) is ( \frac{1}{u} \cdot \frac{du}{dx} ).

Here,

• ( u = \cosh(x) )
• ( \frac{du}{dx} = \sinh(x) )

So,

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

This simplifies to:

$$ \tanh(x) $$

2. Differentiate the second term

Next, we differentiate ( \frac{1}{2\cosh^2(x)} ). We will use the quotient rule ( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} ), where ( f = 1 ) and ( g = 2\cosh^2(x) ).

Calculating the derivatives:

• ( f' = 0 )
• ( g' = 4\cosh(x) \sinh(x) ) (using the chain rule)

The quotient rule yields:

$$ \frac{d}{dx} \left(\frac{1}{2\cosh^2(x)}\right) = \frac{0 \cdot 2\cosh^2(x) - 1 \cdot 4\cosh(x) \sinh(x)}{(2\cosh^2(x))^2} = -\frac{4\cosh(x)\sinh(x)}{4\cosh^4(x)} = -\frac{\sinh(x)}{\cosh^3(x)} $$

3. Combine the derivatives

Now, we combine the results from the two derivative calculations:

$$ \frac{d}{dx}\left(\log(\cosh(x)) + \frac{1}{2\cosh^2(x)}\right) = \tan_h(x) - \frac{\sinh(x)}{\cosh^3(x)} $$

4. Final answer in a simplified form

Now we express the final derivative:

$$ \frac{d}{dx}\left(\log(\cosh(x)) + \frac{1}{2\cosh^2(x)}\right) = \tanh(x) - \frac{\sinh(x)}{\cosh^3(x)} $$