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# How do you find the first and second derivatives of y= (6x^2+4)/ (2x+2) using the quotient rule?

Jun 17, 2018

$y ' = \frac{2 {x}^{2} + 6 x - 2}{x + 1} ^ 2$
$y ' ' = \frac{10}{x + 1} ^ 3$

#### Explanation:

Using the Quotient rule we get

$y ' = \frac{6 x \left(x + 1\right) - \left(3 {x}^{2} + 2\right)}{x + 1} ^ 2$
which simplifyes to

$y ' = \frac{3 {x}^{2} + 6 x - 2}{x + 1} ^ 2$

Again by the Quotient rule we get

$y ' ' = \frac{\left(6 x + 6\right) {\left(x + 1\right)}^{2} - \left(3 {x}^{2} + 6 x - 2\right) \left(2 \left(x + 1\right)\right)}{x + 1} ^ 3$
Simplifying we obtain
$y ' ' = \frac{10}{x + 1} ^ 3$