To find the derivative of the function

$f(x) = \frac{6e^{x}}{6e^{x} + 4}$

we can use the quotient rule, which states that if we have a function $h(x) = \frac{u(x)}{v(x)}$, then the derivative $h'(x)$ is:

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

For this function, we identify:

• $u(x) = 6e^x$
• $v(x) = 6e^x + 4$

Now, we proceed with the steps:

Step 1: Differentiate $u(x)$ and $v(x)$

1. $u(x) = 6e^x \Rightarrow u'(x) = 6e^x$
2. $v(x) = 6e^x + 4 \Rightarrow v'(x) = 6e^x$

Step 2: Apply the Quotient Rule

Now we substitute into the quotient rule formula:

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

Substituting the values we found:

$f'(x) = \frac{(6e^x)(6e^x + 4) - (6e^x)(6e^x)}{(6e^x + 4)^2}$

Step 3: Simplify the Expression

Expanding the terms in the numerator:

$f'(x) = \frac{36e^{2x} + 24e^x - 36e^{2x}}{(6e^x + 4)^2}$

Combine like terms in the numerator:

$f'(x) = \frac{24e^x}{(6e^x + 4)^2}$

So, the derivative of $f(x)$ is:

$f'(x) = \frac{24e^x}{(6e^x + 4)^2}$

Would you like more details, or do you have any questions about this derivative?

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Here are some related questions for further exploration:

1. How would the derivative change if the function were $f(x) = \frac{5e^x}{5e^x + 3}$?
2. What is the second derivative of this function?
3. How can we interpret the derivative of this function in terms of growth and decay?
4. What is the behavior of the derivative as $x \to \infty$ or $x \to -\infty$?
5. How would we find the critical points of $f(x)$?

Tip: Remember that when simplifying derivatives, combining like terms and canceling out can greatly reduce the complexity of the expression.