The problem you’ve uploaded asks to find the limit:

$\lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}$

where $f(x) = \frac{x}{\sqrt{1+x}}$.

This limit is the definition of the derivative of $f(x)$ at $x$. To solve it, we need to differentiate the given function $f(x)$.

Step-by-step solution:

Step 1: Find the derivative of $f(x)$.

We have:

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

Using the quotient rule for derivatives, which states that for $f(x) = \frac{u(x)}{v(x)}$:

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

Here, $u(x) = x$ and $v(x) = \sqrt{1+x}$.

Step 2: Differentiate $u(x)$ and $v(x)$.

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

Step 3: Apply the quotient rule.

$f'(x) = \frac{\sqrt{1+x}(1) - x \left( \frac{1}{2\sqrt{1+x}} \right)}{(1+x)}$

Simplifying:

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

To simplify further, multiply the numerator by $2\sqrt{1+x}$ to combine the terms:

$f'(x) = \frac{2(1+x) - x}{2(1+x)^{3/2}} = \frac{2+x}{2(1+x)^{3/2}}.$

Final Result:

The derivative of $f(x)$ is:

$f'(x) = \frac{2+x}{2(1+x)^{3/2}}.$

Thus, the limit as $h \to 0$ is:

$\lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} = f'(x) = \frac{2+x}{2(1+x)^{3/2}}.$

Let me know if you would like further details or have any questions!

Related Questions:

1. How would the quotient rule change if the function was $\frac{1}{\sqrt{1+x}}$?
2. What is the physical interpretation of the derivative in this problem?
3. How would you find the second derivative of $f(x)$?
4. What happens to the derivative as $x \to \infty$?
5. Can you verify the result using the definition of the derivative directly?

Tip:

Always double-check your use of derivative rules (quotient, product, chain) to avoid small errors when solving calculus problems.