Differentiate the following w.r.t. as indicated :
`(x^(2)-1)^(4//5)" w.r.t. "|x|`

Answer

To differentiate the function \( u = (x^2 - 1)^{\frac{4}{5}} \) with respect to \( v = |x| \), we will use the chain rule. Here are the steps to solve the problem:

### Step-by-Step Solution

1. **Define the Functions**:
Let \( u = (x^2 - 1)^{\frac{4}{5}} \) and \( v = |x| \).

2. **Find \( \frac{du}{dx} \)**:
To find \( \frac{du}{dx} \), we apply the chain rule:
\[ \frac{du}{dx} = \frac{d}{dx} \left( (x^2 - 1)^{\frac{4}{5}} \right) \]
Using the chain rule:
\[ \frac{du}{dx} = \frac{4}{5} (x^2 - 1)^{\frac{4}{5} - 1} \cdot \frac{d}{dx}(x^2 - 1) \]
The derivative of \( x^2 - 1 \) is \( 2x \):
\[ \frac{du}{dx} = \frac{4}{5} (x^2 - 1)^{-\frac{1}{5}} \cdot 2x = \frac{8x}{5(x^2 - 1)^{\frac{1}{5}}} \]

3. **Find \( \frac{dv}{dx} \)**:
The derivative of \( v = |x| \) is:
\[ \frac{dv}{dx} = \frac{x}{|x|} \quad \text{for } x \neq 0 \]

4. **Use the Chain Rule**:
We need to find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} = \frac{du}{dx} \cdot \frac{1}{\frac{dv}{dx}} \]
Thus,
\[ \frac{du}{dv} = \frac{\frac{8x}{5(x^2 - 1)^{\frac{1}{5}}}}{\frac{x}{|x|}} = \frac{8x}{5(x^2 - 1)^{\frac{1}{5}}} \cdot \frac{|x|}{x} \]
For \( x > 0 \), \( |x| = x \) and for \( x < 0 \), \( |x| = -x \). Therefore:
\[ \frac{du}{dv} = \frac{8 |x|}{5(x^2 - 1)^{\frac{1}{5}}} \]

5. **Final Result**:
Thus, the derivative of \( u \) with respect to \( v \) is:
\[ \frac{du}{dv} = \frac{8 |x|}{5 (x^2 - 1)^{\frac{1}{5}}} \]