Given f(x)=(e^(−2x))/(1+e^(10x)); find f′(x).

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.

Given f(x)=(e^(−2x))/(1+e^(10x)); find f′(x).

Find the derivative of f(x) = (e^(-2x)) / (1 + e^(10x)) by applying the quotient rule. This step-by-step solution simplifies the expression, covering key differentiation concepts such as the quotient rule and the exponential function derivative. Suitable for university-level calculus students.

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