# How do you find the first and second derivatives of y = (x^2 + 3) / x using the quotient rule?

Dec 25, 2015

Apply the quotient rule twice to find

$\frac{d}{\mathrm{dx}} \frac{{x}^{2} + 3}{x} = \frac{{x}^{2} - 3}{x} ^ 2$

${d}^{2} / {\mathrm{dx}}^{2} \frac{{x}^{2} + 3}{x} = \frac{6}{x} ^ 3$

#### Explanation:

The Quotient Rule states that

$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} ^ 2 \left(x\right)$

First Derivative:

$\frac{d}{\mathrm{dx}} \frac{{x}^{2} + 3}{x} = \frac{\left(\frac{d}{\mathrm{dx}} \left({x}^{2} + 3\right)\right) x - \left({x}^{2} + 3\right) \left(\frac{d}{\mathrm{dx}} x\right)}{x} ^ 2$

$= \frac{\left(2 x\right) x - \left({x}^{2} + 3\right) \left(1\right)}{x} ^ 2$

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

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

Second Derivative:

${d}^{2} / {\mathrm{dx}}^{2} \frac{{x}^{2} - 3}{x} = \frac{d}{\mathrm{dx}} \frac{{x}^{2} - 3}{x} ^ 2$

$= \frac{\left(\frac{d}{\mathrm{dx}} \left({x}^{2} - 3\right)\right) {x}^{2} - \left({x}^{2} - 3\right) \left(\frac{d}{\mathrm{dx}} {x}^{2}\right)}{{x}^{2}} ^ 2$

$= \frac{\left(2 x\right) {x}^{2} - \left({x}^{2} - 3\right) \left(2 x\right)}{x} ^ 4$

$= \frac{2 {x}^{3} - 2 {x}^{3} + 6 x}{x} ^ 4$

$= \frac{6 x}{x} ^ 4$

$= \frac{6}{x} ^ 3$

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As a side note, the problem could be made slightly easier by initially splitting the function into two terms:

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

$\frac{d}{\mathrm{dx}} \left(1 - \frac{3}{x} ^ 2\right) = - \frac{\left(\frac{d}{\mathrm{dx}} \left(3\right)\right) {x}^{2} - 3 \left(\frac{d}{\mathrm{dx}} {x}^{2}\right)}{{x}^{2}} ^ 2 = - \left(\frac{- 6 x}{x} ^ 4\right) = \frac{6}{x} ^ 3$