Question

Find the derivative of the function. $$ h(x)=\frac{2 x^{2}-3 x+1}{x} $$

Short answer

The derivative of the function \(h(x) = \frac{2x^{2}-3x+1}{x}\) is \(h'(x) = 2 - \frac{1}{x^{2}}\).

Step-by-step solution

Write Down the Function and Quotient Rule

The function is \( h(x)=\frac{2 x^{2}-3 x+1}{x} \). And the Quotient Rule states that the derivative of \(\frac{u}{v}\) is \(\frac{vu' - uv'}{v^{2}}\), where \(u\) and \(v\) are functions of \(x\), and \(u'\) and \(v'\) are their respective derivatives.

Identify u, v, u' and v'

Here, \(u = 2x^2 - 3x + 1\), \(v = x\). So, \(u' = \frac{d}{dx}(2x^2 - 3x + 1) = 4x - 3\), \(v' = \frac{d}{dx}(x) = 1.\)

Apply the Quotient Rule

Substitute \(u\), \(v\), \(u'\), \(v'\) into the Quotient Rule to get: \(h'(x) = \frac{x(4x - 3) - (2x^{2} - 3x + 1)(1)}{x^{2}} = \frac{4x^{2} -3x -2x^2 +3x -1}{x^2} = \frac{2x^{2}-1}{x^2}.\)

Simplify the Result

Simplify the expression: \(h'(x) = \frac{2x^{2}-1}{x^2} = 2 - \frac{1}{x^{2}}\)