Question

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

Answer 1

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

Let's denote:
- \( u = e^x \)
- \( v = \left( \frac{x-2}{x+2} \right)^{3/4} \)

By the product rule, the derivative \( y' \) is given by:

\[
y' = u'v + uv'
\]

**Step 1: Differentiate \( u \)**

Since \( u = e^x \):
\[
u' = e^x
\]

**Step 2: Differentiate \( v \)**

To differentiate \( v = \left( \frac{x-2}{x+2} \right)^{3/4} \), we apply the chain rule. Let:
\[
w = \frac{x-2}{x+2}
\]
so \( v = w^{3/4} \).

First, find \( w' \):
\[
w = \frac{x-2}{x+2}
\]

Using the quotient rule:
\[
w' = \frac{(x+2)(1) - (x-2)(1)}{(x+2)^2} = \frac{x + 2 - x + 2}{(x+2)^2} = \frac{4}{(x+2)^2}
\]

Now differentiate \( v \):
\[
v' = \frac{3}{4} w^{-\frac{1}{4}} w' = \frac{3}{4} \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}} \cdot \frac{4}{(x+2)^2}
\]

This simplifies to:
\[
v' = 3 \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}} \cdot \frac{1}{(x+2)^2}
\]

**Step 3: Substitute back into the product rule**

Now substituting \( u', v, u, \) and \( v' \) back into the product rule:

\[
y' = e^x \left( \frac{x-2}{x+2} \right)^{3/4} + e^x \cdot 3 \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}} \cdot \frac{1}{(x+2)^2}
\]

Thus, we have the complete derivative:
\[
y' = e^x \left( \frac{x-2}{x+2} \right)^{3/4} + \frac{3 e^x}{(x+2)^2} \left( \frac{x-2}{x+2} \right)^{-\frac{1}{4}}
\]

This is the derivative of the given function \( y \).