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 of the function $$ y = \frac{1 + x}{1 - x} $$ with respect to $$ x $$, we can use the **quotient rule**. The quotient rule states that if you have a function $$ y = \frac{u(x)}{v(x)} $$, then its derivative is:

$$
\frac{dy}{dx} = \frac{v(x) \cdot \frac{du}{dx} - u(x) \cdot \frac{dv}{dx}}{[v(x)]^2}
$$

Applying this to your function:

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

2. **Compute the derivatives $$ \frac{du}{dx} $$ and $$ \frac{dv}{dx} $$:**
   $$
   \frac{du}{dx} = 1 \quad \text{and} \quad \frac{dv}{dx} = -1
   $$

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

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

5. **Write the final expression for the 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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