If $$ y=\frac{1+x}{1-x} $$, find $$ \frac{d y}{d x} $$

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UpStudy AI · Accepted Answer

To find the derivative $$ \frac{dy}{dx} $$ where $$ y = \frac{1+x}{1-x} $$, you can use the **quotient rule** for differentiation. The quotient rule states that if you have a function $$ y = \frac{u}{v} $$, then the derivative $$ y' $$ is given by:

$$
y' = \frac{u'v - uv'}{v^2}
$$

Applying this to $$ y = \frac{1+x}{1-x} $$:

1. **Identify $$ u $$ and $$ v $$:**
   $$
   u = 1 + x \quad \text{and} \quad v = 1 - x
   $$

2. **Find the derivatives $$ u' $$ and $$ v' $$:**
   $$
   u' = \frac{d}{dx}(1 + x) = 1
   $$
   $$
   v' = \frac{d}{dx}(1 - x) = -1
   $$

3. **Apply the quotient rule:**
   $$
   \frac{dy}{dx} = \frac{(1)(1 - x) - (1 + x)(-1)}{(1 - x)^2}
   $$

4. **Simplify the numerator:**
   $$
   (1)(1 - x) - (1 + x)(-1) = (1 - x) + (1 + x) = 2
   $$

5. **Write the final derivative:**
   $$
   \frac{dy}{dx} = \frac{2}{(1 - x)^2}
   $$

**Final Answer:**
$$
\frac{dy}{dx} = \frac{2}{\,\left(1 - x\,\right)^2\,}
$$
$$ \frac{dy}{dx} = \frac{2}{(1 - x)^2} $$

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