To differentiate the function $f(x) = \frac{e^x}{2x + 1}$, we'll use the quotient rule for derivatives.

The quotient rule states that if you have a function of the form:

$f(x) = \frac{g(x)}{h(x)}$

Then its derivative is given by:

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

Here, $g(x) = e^x$ and $h(x) = 2x + 1$.

Step 1: Differentiate $g(x)$ and $h(x)$

• The derivative of $g(x) = e^x$ is:
  $g'(x) = e^x$

• The derivative of $h(x) = 2x + 1$ is:
  $h'(x) = 2$

Step 2: Apply the Quotient Rule

Now, we can apply the quotient rule:

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

Simplifying the numerator:

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

Final Answer:

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

Would you like more details on this solution, or do you have any questions?

Related Questions:

1. How do you apply the product rule for derivatives?
2. What is the chain rule, and how does it differ from the quotient rule?
3. How would you integrate a function like $\frac{e^x}{2x + 1}$?
4. Can the quotient rule be applied when $h(x)$ is a constant?
5. What are higher-order derivatives, and how are they calculated for this function?

Tip: When using the quotient rule, always remember to simplify the expression as much as possible before finalizing the answer!