# How do you find the first and second derivatives of y= (x^2 + 2x + 5) / (x + 1) using the quotient rule?

Feb 19, 2018

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

Second derivative: $\frac{8}{x + 1} ^ 3$

#### Explanation:

The quotient rule is:

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

And in our problem, $f$ is the numerator, and $g$ is the denominator of $\frac{{x}^{2} + 2 x + 5}{x + 1}$.

Using the rule:

$\left(\frac{{x}^{2} + 2 x + 5}{x + 1}\right) ' = \frac{\left({x}^{2} + 2 x + 5\right) ' \left(x + 1\right) - \left({x}^{2} + 2 x + 5\right) \left(x + 1\right) '}{x + 1} ^ 2$

$= \frac{\left(2 x + 2\right) \left(x + 1\right) - \left({x}^{2} + 2 x + 5\right) \left(1\right)}{x + 1} ^ 2$

$= \frac{2 {x}^{2} + 4 x + 2 - {x}^{2} - 2 x - 5}{x + 1} ^ 2$

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

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

That's the first derivative. To find the next one, perform the same actions, but with this new fraction:

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

After a lot of simplifying:

$\frac{8}{x + 1} ^ 3$

Feb 19, 2018

first derivative: $\frac{\left(x + 3\right) \left(x - 1\right)}{x + 1} ^ 2$
second derivative: $\frac{8}{x + 1} ^ 3$

#### Explanation:

$\left(\frac{u}{v}\right) ' = \frac{u ' v - v ' u}{{v}^{2}}$

$u \left(x\right) = {x}^{2} + 2 x + 5$

power rule: $\left({x}^{n}\right) ' = n {x}^{n - 1}$

$u \left(x\right) = {x}^{2} + 2 x + 5$

$u ' \left(x\right) = 2 {x}^{1} + 2 {x}^{0} = 2 x + 2$

$v \left(x\right) = x + 1$

$v ' \left(x\right) = 1 {x}^{0} = 1$

$u ' v = \left(2 x + 2\right) \cdot \left(x + 1\right) = 2 {x}^{2} + 4 x + 2$

$v ' u = 1 \cdot \left({x}^{2} + 2 x + 5\right) = {x}^{2} + 2 x + 5$

${v}^{2} = {\left(x + 1\right)}^{2}$

$\left(u ' v - v ' u\right) = \left(2 {x}^{2} + 4 x + 2\right) - \left({x}^{2} + 2 x + 5\right)$

$= 2 {x}^{2} + 4 x + 2 - {x}^{2} - 2 x - 5$

$= {x}^{2} + 2 x - 3$

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

the first derivative of $y = \frac{{x}^{2} + 2 x + 5}{x + 1}$ is $\frac{{x}^{2} + 2 x - 3}{x + 1} ^ 2$

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the quotient rule $\left(\frac{u}{v}\right) ' = \frac{u ' v - v ' u}{{v}^{2}}$ can be used again

where $u \left(x\right) = {x}^{2} + 2 x - 3$

and $v \left(x\right) = {\left(x + 1\right)}^{2} , \mathmr{and} {x}^{2} + 2 x + 1$.

$u ' \left(x\right) = 2 {x}^{1} + 2 {x}^{0} = 2 x + 2 , \mathmr{and} 2 \left(x + 1\right)$

$v ' \left(x\right) = 2 {x}^{1} + 2 {x}^{0} = 2 x + 2 , \mathmr{and} 2 \left(x + 1\right)$

$u ' v = \left(2 \left(x + 1\right)\right) \cdot {\left(x + 1\right)}^{2} = 2 {\left(x + 1\right)}^{3} = 2 {x}^{3} + 6 {x}^{2} + 6 x + 2$

$v ' u = \left(2 x + 2\right) \cdot \left({x}^{2} + 2 x - 3\right) = 2 {x}^{3} + 2 {x}^{2} + 4 {x}^{2} + 4 x - 6 x - 6$

$= 2 {x}^{3} + 6 {x}^{2} - 2 x - 6$

$u ' v - v ' u = \left(2 {x}^{3} + 6 {x}^{2} + 6 x + 2\right) - \left(2 {x}^{3} + 6 {x}^{2} - 2 x - 6\right)$

$= 2 {x}^{3} + 6 {x}^{2} + 6 x + 2 - 2 {x}^{3} - 6 {x}^{2} + 2 x + 6$

$= 8 x + 8$

(u'v - v'u)/v^2 = (8x + 8)/((x+1)^4

$8 x + 8 = 8 \left(x + 1\right)$

$\frac{8 x + 8}{x + 1} ^ 4 = \frac{8 \left(x + 1\right)}{x + 1} ^ 4$

$= \frac{8}{x + 1} ^ 3$

the second derivative of $y = \frac{{x}^{2} + 2 x + 5}{x + 1}$ is $\frac{8}{x + 1} ^ 3$