To find the derivative of $\frac{x+1}{\sqrt{x}}$ using the quotient rule, we proceed as follows.

The quotient rule states that if you have a function of the form $\frac{f(x)}{g(x)}$, the derivative is:

$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$

For the given function $\frac{x+1}{\sqrt{x}}$:

• $f(x) = x+1$
• $g(x) = \sqrt{x} = x^{1/2}$

Now, calculate the derivatives of $f(x)$ and $g(x)$:

1. $f'(x) = \frac{d}{dx}(x+1) = 1$
2. $g'(x) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Now, apply the quotient rule:

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

Simplify step by step:

1. The denominator is: $(\sqrt{x})^2 = x$

2. For the numerator: $\sqrt{x} \cdot 1 = \sqrt{x}$ $(x+1) \cdot \frac{1}{2\sqrt{x}} = \frac{x+1}{2\sqrt{x}}$

Thus, the derivative becomes: $\frac{\sqrt{x} - \frac{x+1}{2\sqrt{x}}}{x}$

To simplify the numerator further, combine the terms by finding a common denominator:

$\frac{\sqrt{x}}{1} - \frac{x+1}{2\sqrt{x}} = \frac{2\sqrt{x}}{2} - \frac{x+1}{2\sqrt{x}} = \frac{2\sqrt{x} - (x+1)}{2\sqrt{x}}$

Now, the expression for the derivative is: $\frac{\frac{2\sqrt{x} - (x+1)}{2\sqrt{x}}}{x} = \frac{2\sqrt{x} - (x+1)}{2x\sqrt{x}}$

Thus, the derivative of $\frac{x+1}{\sqrt{x}}$ is:

$\frac{2\sqrt{x} - (x+1)}{2x\sqrt{x}}$

Let me know if you'd like more details or clarifications!

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Here are five related questions to explore further:

1. What is the derivative of $\frac{2x+3}{\sqrt{x}}$ using the quotient rule?
2. How would you differentiate $\frac{1}{x+1}$ using the quotient rule?
3. What is the significance of the quotient rule when applied to rational functions?
4. Can the chain rule be used alongside the quotient rule for more complex expressions?
5. How would you apply the quotient rule to a function involving both sine and cosine?

Tip: When dealing with rational functions like $\frac{f(x)}{g(x)}$, always simplify the result as much as possible after applying the quotient rule.