It seems that you are working on a problem involving derivatives and division of functions. Based on the image you uploaded, the equation shown is:

$g(x) = \frac{\sqrt{5x+1}}{\sqrt{x}}$

To solve the derivative of this function using the quotient rule, we recall that the quotient rule states:

$\left(\frac{f(x)}{h(x)}\right)' = \frac{f'(x)h(x) - f(x)h'(x)}{(h(x))^2}$

In this case:

• $f(x) = \sqrt{5x + 1}$
• $h(x) = \sqrt{x}$

Step-by-Step Solution:

1. Find the derivatives of $f(x)$ and $h(x)$:

   • Derivative of $f(x) = \sqrt{5x+1}$: $f'(x) = \frac{d}{dx} \left( (5x + 1)^{1/2} \right) = \frac{1}{2} (5x + 1)^{-1/2} \times 5 = \frac{5}{2\sqrt{5x + 1}}$

   • Derivative of $h(x) = \sqrt{x}$: $h'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$

2. Apply the quotient rule: $g'(x) = \frac{\left(\frac{5}{2\sqrt{5x + 1}}\right) \cdot \sqrt{x} - \left(\sqrt{5x + 1}\right) \cdot \frac{1}{2\sqrt{x}}}{(\sqrt{x})^2}$

   Simplifying: $g'(x) = \frac{\frac{5\sqrt{x}}{2\sqrt{5x + 1}} - \frac{\sqrt{5x + 1}}{2\sqrt{x}}}{x}$

3. Combine the terms under the common denominator: $g'(x) = \frac{1}{2x} \left( \frac{5\sqrt{x}}{\sqrt{5x + 1}} - \frac{\sqrt{5x + 1}}{\sqrt{x}} \right)$

This is the derivative of the function $g(x)$.

Let me know if you need more details or further clarification on this solution.

Related Questions:

1. How would you simplify the expression for $g'(x)$?
2. What is the behavior of the function $g(x)$ as $x \to 0$?
3. How would you apply the quotient rule to functions involving higher powers?
4. What happens when you apply the chain rule to composite functions in the quotient rule?
5. Can you find the second derivative of $g(x)$?

Tip:

When dealing with functions that are quotients, always check if simplifications can be made before applying the quotient rule to make the problem easier.