To find the derivative of the function given by

$f(x) = \frac{1}{\sqrt{1 + x^2} \cdot \left( x + \sqrt{1 + x^2} \right)},$

we'll use the quotient rule for derivatives, which states that if $f(x) = \frac{g(x)}{h(x)}$, then

$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.$

In this case:

• $g(x) = 1$ (constant),
• $h(x) = \sqrt{1 + x^2} \cdot (x + \sqrt{1 + x^2})$.

Step-by-Step Solution

1. Differentiate $g(x)$: Since $g(x) = 1$, $g'(x) = 0$.

2. Differentiate $h(x)$: We need to apply the product rule here because $h(x) = \sqrt{1 + x^2} \cdot (x + \sqrt{1 + x^2})$.

   Let $u(x) = \sqrt{1 + x^2}$ and $v(x) = x + \sqrt{1 + x^2}$. Then $h(x) = u(x) \cdot v(x)$, and the derivative is: $h'(x) = u'(x)v(x) + u(x)v'(x).$

   • To find $u'(x)$, note that $u(x) = (1 + x^2)^{1/2}$, so using the chain rule, $u'(x) = \frac{1}{2}(1 + x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}.$

   • For $v(x) = x + \sqrt{1 + x^2}$, we find $v'(x)$ as follows: $v'(x) = 1 + \frac{x}{\sqrt{1 + x^2}}.$

   Substituting $u'(x)$, $u(x)$, $v(x)$, and $v'(x)$ into the expression for $h'(x)$: $h'(x) = \frac{x}{\sqrt{1 + x^2}} \cdot \left( x + \sqrt{1 + x^2} \right) + \sqrt{1 + x^2} \cdot \left( 1 + \frac{x}{\sqrt{1 + x^2}} \right).$

3. Compute $f'(x)$: Since $f(x) = \frac{g(x)}{h(x)} = \frac{1}{h(x)}$, we have [ f'(x) = \frac{0 \cdot h(x) - 1 \cdot h'(x)}{[h(x)]^2} = -\frac{h'(x)}{[h(x)]^2}. ]

So, the derivative $f'(x)$ is

$f'(x) = -\frac{h'(x)}{[h(x)]^2}.$

Would you like a more detailed simplification of $h'(x)$ or further steps on this derivative?

1. What would be the next steps to simplify $f'(x)$ further?
2. How does the product rule apply when differentiating composite functions like this?
3. What are common mistakes to avoid in applying the quotient rule?
4. How does the chain rule factor into calculating $u'(x)$ and $v'(x)$?
5. How could this derivative function be used in practical applications?

Tip: Always simplify each part of the derivative before combining terms to reduce the chance of error in complex expressions.