We are tasked with finding the derivative of the function:

$f(x) = \sqrt{\frac{x^2 + 2x + 3}{\ln(x)}}$

Step 1: Rewriting the function

We rewrite the square root as a fractional exponent: $f(x) = \left(\frac{x^2 + 2x + 3}{\ln(x)}\right)^{\frac{1}{2}}$

Step 2: Apply the chain rule

Using the chain rule, the derivative of $f(x)$ is: $f'(x) = \frac{1}{2} \left(\frac{x^2 + 2x + 3}{\ln(x)}\right)^{-\frac{1}{2}} \cdot \frac{d}{dx} \left( \frac{x^2 + 2x + 3}{\ln(x)} \right)$

This simplifies to: $f'(x) = \frac{1}{2\sqrt{\frac{x^2 + 2x + 3}{\ln(x)}}} \cdot \frac{d}{dx} \left( \frac{x^2 + 2x + 3}{\ln(x)} \right)$

Step 3: Differentiate the inner fraction

Let $u(x) = x^2 + 2x + 3$ and $v(x) = \ln(x)$. Then the fraction can be written as $\frac{u(x)}{v(x)}$. Using the quotient rule: $\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

Compute $u'(x)$ and $v'(x)$:

$u'(x) = 2x + 2, \quad v(x) = \ln(x), \quad v'(x) = \frac{1}{x}$

Substitute into the quotient rule:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{(2x + 2)\ln(x) - (x^2 + 2x + 3)\frac{1}{x}}{(\ln(x))^2}$

Simplify the terms:

= \frac{(2x + 2)\ln(x) - (x + 2 + \frac{3}{x})}{(\ln(x))^2}$$ ### Step 4: Final expression for $$f'(x)$$ Substitute this result back into the derivative of $$f(x)$$: $$f'(x) = \frac{1}{2\sqrt{\frac{x^2 + 2x + 3}{\ln(x)}}} \cdot \frac{(2x + 2)\ln(x) - (x + 2 + \frac{3}{x})}{(\ln(x))^2}$$ This is the derivative of the given function. It can be simplified further if needed. Would you like me to work through the simplification or provide an alternate explanation? --- ### Relative Questions: 1. What is the chain rule and how is it applied in derivatives? 2. How does the quotient rule help in differentiating fractions? 3. What happens if $$\ln(x)$$ is very small or approaches zero? 4. How can we simplify the derivative expression further? 5. How does the square root affect the differentiation process? ### Tip: When differentiating composite functions, always identify the outermost function first and apply the chain rule systematically.