To find the derivative $f'(x)$ of the function $f(x) = \frac{e^{-2x}}{1 + e^{10x}}$, we'll need to apply the quotient rule for differentiation, which is stated as follows:

$\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

In this case:

• $u(x) = e^{-2x}$
• $v(x) = 1 + e^{10x}$

Step 1: Find $u'(x)$

The derivative of $u(x) = e^{-2x}$ is: $u'(x) = -2e^{-2x}$

Step 2: Find $v'(x)$

The derivative of $v(x) = 1 + e^{10x}$ is: $v'(x) = 10e^{10x}$

Step 3: Apply the quotient rule

Now apply the quotient rule: $f'(x) = \frac{(-2e^{-2x})(1 + e^{10x}) - (e^{-2x})(10e^{10x})}{(1 + e^{10x})^2}$

Step 4: Simplify the expression

First, distribute the terms in the numerator: $f'(x) = \frac{-2e^{-2x}(1 + e^{10x}) - 10e^{-2x}e^{10x}}{(1 + e^{10x})^2}$

Simplify further: $f'(x) = \frac{-2e^{-2x} - 2e^{-2x}e^{10x} - 10e^{8x}}{(1 + e^{10x})^2}$

Final Answer:

$f'(x) = \frac{-2e^{-2x} - 12e^{8x}}{(1 + e^{10x})^2}$

This is the derivative of the given function.

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Would you like more details on any step or further explanation?

Here are 5 related questions to expand on this topic:

1. How does the quotient rule differ from the product rule?
2. Can you explain why the derivative of $e^{ax}$ is $ae^{ax}$?
3. How can we apply the chain rule within the quotient rule?
4. What are some common mistakes when applying the quotient rule?
5. Can you find the second derivative of $f(x)$?

Tip: Always simplify the numerator after applying the quotient rule to avoid unnecessary complexity in your expressions.