Differentiate (cosh(2x)/x) + (√x * sinh(2x)) with respect to x.

Steps to Solve

1. Identify the function to differentiate

We have the function ( f(x) = \frac{\cosh(2x)}{x} + \sqrt{x} \cdot \sinh(2x} ).

2. Differentiate the first term using the quotient rule

For the term ( \frac{\cosh(2x)}{x} ), we apply the quotient rule, which states that if ( u(x) = \cosh(2x) ) and ( v(x) = x ), then

$$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} $$

First, we calculate ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):

• ( \frac{du}{dx} = 2\sinh(2x) ) (using the chain rule)
• ( \frac{dv}{dx} = 1 )

Now substituting into the quotient rule:

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

3. Differentiate the second term using the product rule

For the term ( \sqrt{x} \cdot \sinh(2x) ), we use the product rule, which states that

$$ \frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx} $$

Let ( u(x) = \sqrt{x} ) and ( v(x) = \sinh(2x) ).

Here, we compute ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):

• ( \frac{du}{dx} = \frac{1}{2\sqrt{x}} )
• ( \frac{dv}{dx} = 2\cosh(2x) )

Now applying the product rule:

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

4. Combine the derivatives

Now we combine the derivatives of both terms:

$$ f'(x) = \frac{x(2\sinh(2x)) - \cosh(2x)}{x^2} + \sqrt{x}(2\cosh(2x)) + \frac{\sinh(2x)}{2\sqrt{x}} $$

5. Simplify the expression if necessary

While further simplification may involve combining fractions or simplifying terms, this already gives us the form of the derivative.