Differentiate √x cosech(√x), x>0 with respect to x.

Steps to Solve

1. Identify the function for differentiation

The function we want to differentiate is $f(x) = \sqrt{x} \cdot \text{cosech}(\sqrt{x})$.

2. Apply the product rule

We need to use the product rule for differentiation, which states that if you have a function $u(x) \cdot v(x)$, then the derivative is given by:

$$ (u \cdot v)' = u' \cdot v + u \cdot v' $$

Here, let $u(x) = \sqrt{x}$ and $v(x) = \text{cosech}(\sqrt{x})$.

3. Differentiate each part

First, we find the derivative of $u(x)$:

$$ u'(x) = \frac{1}{2\sqrt{x}} $$

Next, we need to differentiate $v(x)$. The derivative of $\text{cosech}(x)$ is $-\text{cosech}(x) \cdot \text{coth}(x)$.

Using the chain rule, we have:

$$ v'(x) = -\text{cosech}(\sqrt{x}) \cdot \text{coth}(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) = -\text{cosech}(\sqrt{x}) \cdot \text{coth}(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} $$

4. Combine the derivatives using the product rule

Now we will substitute $u(x)$, $u'(x)$, $v(x)$, and $v'(x)$ back into the product rule:

$$ f'(x) = u' \cdot v + u \cdot v' = \left(\frac{1}{2\sqrt{x}}\right) \cdot \text{cosech}(\sqrt{x}) + \sqrt{x} \cdot \left(-\text{cosech}(\sqrt{x}) \cdot \text{coth}(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\right) $$

5. Simplify the expression

We can simplify the expression:

$$ f'(x) = \frac{1}{2\sqrt{x}} \cdot \text{cosech}(\sqrt{x}) - \frac{1}{2} \cdot \text{cosech}(\sqrt{x}) \cdot \text{coth}(\sqrt{x}) $$

6. Final expression for the derivative

Thus, we arrive at the derivative of the original function as:

$$ f'(x) = \frac{1}{2\sqrt{x}} \cdot \text{cosech}(\sqrt{x}) - \frac{1}{2} \cdot \text{cosech}(\sqrt{x}) \cdot \text{coth}(\sqrt{x}) $$