91Ó°ÊÓ

In order to differentiate \(h(x) = \frac{2x^{2}-3x+1}{x}\), the quadratic with the coefficients can be thought of being equal to \(g(x)\) and \(x\) being the \(h(x)\). Now, we firstly need to find the derivative of both \(g(x)\) and \(h(x)\) denoted as \(g'(x)\) and \(h'(x)\). \n\n According to the power rule, the derivative of \(2x^2\) is \(4x\), the derivative of \(-3x\) is \(-3\) and the derivative of the constant \(1\) is \(0\). Adding all these results up, we find that \(g'(x) = 4x -3\). \n\n As for \(h'(x)\), since \(h(x) = x\), the derivative is \(1\) due to the power rule as well. \n\n The next step is to apply the quotient rule. Substituting the original \(g(x)\), \(h(x)\), \(g'(x)\) and \(h'(x)\) into the quotient rule yields \(\frac{x*(4x -3) - (2x^2 -3x+1) * 1}{x^2}\).

After substituting the values into the quotient rule equation, expand the numerator and then simplify it: \n\nThe numerator turns out to be of the form \(x*4x -3x-(2x^2 -3x+1) = 4x^2 -3x -(2x^2 -3x+1) = 2x^2 - 1\). The simplification was done by multiplying out, combining like terms, and subtracting across.

Substituting the simplified numerator back into our derivative function, we get the final derivative of the function: \(h'(x) = \frac{2x^2 - 1}{x^2}\). That's the final result.