The given function is:

$f(x) = \left( \frac{1 + x}{1 - x} \right)^{10}, \, x = 0$

To find the derivative $f'(x)$ at $x = 0$, we will proceed step by step:

Step 1: Rewrite the function

Let: $g(x) = \frac{1 + x}{1 - x}$ So, $f(x) = g(x)^{10}$

Step 2: Differentiate using the chain rule

Using the chain rule: $f'(x) = 10 \cdot g(x)^9 \cdot g'(x)$

Now, we need to compute $g'(x)$.

Step 3: Differentiate $g(x) = \frac{1 + x}{1 - x}$

Using the quotient rule: $g'(x) = \frac{(1 - x) \cdot (1) - (1 + x) \cdot (-1)}{(1 - x)^2}$ Simplify: $g'(x) = \frac{(1 - x) + (1 + x)}{(1 - x)^2} = \frac{2}{(1 - x)^2}$

Step 4: Substitute into $f'(x)$

Substituting $g(x)$ and $g'(x)$ into $f'(x)$: $f'(x) = 10 \cdot \left( \frac{1 + x}{1 - x} \right)^9 \cdot \frac{2}{(1 - x)^2}$

Step 5: Evaluate at $x = 0$

At $x = 0$: $g(0) = \frac{1 + 0}{1 - 0} = 1, \quad g'(0) = \frac{2}{(1 - 0)^2} = 2$

Thus: $f'(0) = 10 \cdot (1)^9 \cdot 2 = 10 \cdot 2 = 20$

Final Answer:

$f'(0) = 20$

Would you like a detailed explanation of any step or further help? 😊

5 Related Questions:

1. How do we compute higher-order derivatives of $f(x)$ at $x = 0$?
2. What is the derivative of $g(x)$ for general $x$ without substitution?
3. How does the chain rule apply to composite functions like $f(x)$?
4. How can we evaluate $f'(x)$ symbolically for other values of $x$?
5. Why does the quotient rule give $g'(x) = \frac{2}{(1-x)^2}$?

Tip:

When differentiating powers of functions, always apply the chain rule carefully to avoid errors.