Find the derivative of :
`(x^4+3x^3+4x^2+2)/(x^3)`

Answer

To find the derivative of the function \( f(x) = \frac{x^4 + 3x^3 + 4x^2 + 2}{x^3} \), we can follow these steps:

### Step 1: Simplify the function
First, we can simplify the function by dividing each term in the numerator by \( x^3 \):

\[ f(x) = \frac{x^4}{x^3} + \frac{3x^3}{x^3} + \frac{4x^2}{x^3} + \frac{2}{x^3} \]

This simplifies to:

\[ f(x) = x + 3 + \frac{4}{x} + \frac{2}{x^3} \]

### Step 2: Rewrite the function using negative exponents
Next, we can rewrite the terms with \( x \) in the denominator using negative exponents:

\[ f(x) = x + 3 + 4x^{-1} + 2x^{-3} \]

### Step 3: Differentiate the function
Now we can differentiate \( f(x) \) using the power rule, which states that the derivative of \( x^n \) is \( n x^{n-1} \):

1. The derivative of \( x \) is \( 1 \).
2. The derivative of \( 3 \) (a constant) is \( 0 \).
3. The derivative of \( 4x^{-1} \) is \( -4x^{-2} \) (using the power rule).
4. The derivative of \( 2x^{-3} \) is \( -6x^{-4} \) (again using the power rule).

Putting it all together, we have:

\[ f'(x) = 1 + 0 - 4x^{-2} - 6x^{-4} \]

### Step 4: Simplify the derivative
We can simplify the derivative to:

\[ f'(x) = 1 - \frac{4}{x^2} - \frac{6}{x^4} \]

### Final Answer
Thus, the derivative of the function is:

\[ f'(x) = 1 - \frac{4}{x^2} - \frac{6}{x^4} \]