- #1

caters

- 229

- 9

## Homework Statement

if $$y = \frac{2x^5-3x^3+x^2}{x^3}$$ then $$\frac{dy}{dx} =$$

## Homework Equations

if $$f(x) = x^n$$ then $$f'(x) = nx^{n-1}$$

## The Attempt at a Solution

$$\frac{2x^5-3x^3+x^2}{x^3} = \frac{2x^5}{x^3} - \frac{3x^3}{x^3} + \frac{x^2}{x^3}$$

$$ f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - \frac{9x^2}{3x^2} + \frac{2x}{3x^2}$$

$$ f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - 3 + \frac{2x}{3x^2}$$

And now I am stuck as to how to simplify this. Should I have done the whole polynomial division before I took the derivative?