Question

Differentiate e^x( (x-2/x+2)^3/4)

Answer 1

To differentiate the function \( y = e^x \left( \frac{x-2}{x+2} \right)^{\frac{3}{4}} \), we will use the product rule and the chain rule.

1. **Identify the parts of the function:**
Let
\[
u = e^x \quad \text{and} \quad v = \left( \frac{x-2}{x+2} \right)^{\frac{3}{4}}.
\]

2. **Differentiate \( u \) and \( v \):**
- For \( u \):
\[
u' = e^x.
\]
- For \( v \):
Using chain rule:
\[
v = \left( \frac{x-2}{x+2} \right)^{\frac{3}{4}},
\]
let \( z = \frac{x-2}{x+2} \). Then,
\[
v = z^{\frac{3}{4}}.
\]

We need \( \frac{dz}{dx} \):
\[
z = \frac{x-2}{x+2}.
\]
Using the quotient rule:
\[
\frac{dz}{dx} = \frac{(x+2)(1) - (x-2)(1)}{(x+2)^2} = \frac{(x+2) - (x-2)}{(x+2)^2} = \frac{4}{(x+2)^2}.
\]
Now, using the chain rule to differentiate \( v \):
\[
v' = \frac{3}{4} z^{-\frac{1}{4}} \cdot \frac{dz}{dx} = \frac{3}{4} \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}} \cdot \frac{4}{(x+2)^2} = \frac{3}{(x+2)^2} \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}}.
\]

3. **Combine using product rule:**
Using the product rule \( y' = u'v + uv' \):
\[
y' = e^x v + e^x v'.
\]
Substituting \( u \) and \( v \):
\[
y' = e^x \left( \frac{x-2}{x+2} \right)^{\frac{3}{4}} + e^x \cdot \frac{3}{(x+2)^2} \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}}.
\]

4. **Factor out common terms:**
\[
y' = e^x \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}} \left( \left( \frac{x-2}{x+2} \right) + \frac{3}{(x+2)^2} \right).
\]

5. **Final expression:**
The derivative of \( y \) is:
\[
y' = e^x \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}} \left( \frac{x-2}{x+2} + \frac{3}{(x+2)^2} \right).
\]