Question
If \( y=\frac{1+x}{1-x} \), find \( \frac{d y}{d x} \)
Ask by Mcdonald Weber. in Pakistan
Jan 20,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\[
\frac{dy}{dx} = \frac{2}{(1 - x)^2}
\]
Solution
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:**
\[
\frac{dy}{dx} = \frac{2}{\,\left(1 - x\,\right)^2\,}
\]
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Simplify this solution Bonus Knowledge
To find the derivative \( \frac{d y}{d x} \) of the function \( y = \frac{1+x}{1-x} \), we can use the quotient rule, which states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \). Here, let \( u = 1+x \) and \( v = 1-x \). First, we compute the derivatives of \( u \) and \( v \): - \( \frac{du}{dx} = 1 \) - \( \frac{dv}{dx} = -1 \) Now, applying the quotient rule: \[ \frac{d y}{d x} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} \] Simplifying the numerator: \[ (1-x) + (1+x) = 1 - x + 1 + x = 2 \] Thus, \[ \frac{d y}{d x} = \frac{2}{(1-x)^2} \] In conclusion, the derivative is: \[ \frac{d y}{d x} = \frac{2}{(1-x)^2} \]
