Find `dy/dx` for each of the following
`y=(1/(1+x))(x^(-2)+2/x-1)+root(3)(x)-1/(root(3)(x))`

Answer

To find \( \frac{dy}{dx} \) for the function

\[ y = \left(\frac{1}{1+x}\right)\left(x^{-2} + \frac{2}{x} - 1\right) + \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \]

we will differentiate it step by step.

### Step 1: Rewrite the function

First, let's rewrite the function in a more manageable form:

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

### Step 2: Differentiate the first term using the product rule

Let

\[ u = \frac{1}{1+x} \quad \text{and} \quad v = x^{-2} + 2x^{-1} - 1 \]

Using the product rule \( \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \):

1. Differentiate \( u \):

\[ \frac{du}{dx} = -\frac{1}{(1+x)^2} \]

2. Differentiate \( v \):

\[ \frac{dv}{dx} = -2x^{-3} - 2x^{-2} \]

Now, applying the product rule:

\[ \frac{d}{dx} \left( \frac{1}{1+x} \left( x^{-2} + 2x^{-1} - 1 \right) \right) = \frac{1}{1+x} \left( -2x^{-3} - 2x^{-2} \right) + \left( x^{-2} + 2x^{-1} - 1 \right) \left( -\frac{1}{(1+x)^2} \right) \]

### Step 3: Differentiate the second term

Now, differentiate the second part of \( y \):

\[ \frac{d}{dx} \left( x^{1/3} - x^{-1/3} \right) = \frac{1}{3} x^{-2/3} + \frac{1}{3} x^{-4/3} \]

### Step 4: Combine the derivatives

Now we combine all the derivatives:

\[ \frac{dy}{dx} = \frac{1}{1+x} \left( -2x^{-3} - 2x^{-2} \right) - \left( x^{-2} + 2x^{-1} - 1 \right) \frac{1}{(1+x)^2} + \frac{1}{3} x^{-2/3} + \frac{1}{3} x^{-4/3} \]

### Step 5: Simplify the expression

Combine and simplify the expression as needed.

### Final Answer

The final expression for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \text{(combined and simplified terms)} \]