To find the derivative of the function $f(x) = \frac{x}{\sqrt{x+1}}$, we can use the Quotient Rule, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'v - uv'}{v^2}$.

In this case, we define:

• $u(x) = x$
• $v(x) = \sqrt{x+1}$

Step 1: Find $u'$ and $v'$

1. $u(x) = x$ so $u' = 1$.
2. $v(x) = \sqrt{x+1} = (x+1)^{1/2}$. To differentiate, we use the chain rule: $v' = \frac{1}{2}(x+1)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x+1}}.$

Step 2: Apply the Quotient Rule

Now, substitute $u$, $u'$, $v$, and $v'$ into the Quotient Rule formula: $f'(x) = \frac{u'v - uv'}{v^2}.$

Substitute the values: $f'(x) = \frac{(1) \cdot \sqrt{x+1} - x \cdot \frac{1}{2\sqrt{x+1}}}{(\sqrt{x+1})^2}.$

Step 3: Simplify the Expression

1. The denominator simplifies as follows: $(\sqrt{x+1})^2 = x + 1.$

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

So the derivative is: $f'(x) = \frac{x+2}{2(x+1)^{3/2}}.$

Would you like further details or clarification on any steps?

Related Questions

1. How can we verify this derivative using a different method?
2. What is the second derivative of $f(x)$?
3. How does this function behave as $x$ approaches infinity?
4. Can we find the critical points of $f(x)$?
5. What is the behavior of $f(x)$ as $x$ approaches -1?

Tip

When using the Quotient Rule, it helps to simplify each derivative term separately before combining them.