Question

if $f(x)=\frac{x^{2}+3x + 2}{x + 3}$, then $f(x)=$
(a) $2x+3$
(b) $\frac{-x^{2}-6x - 7}{(x + 3)^{2}}$
(c) $\frac{x^{2}+6x + 7}{(x + 3)^{2}}$
(d) $\frac{x^{2}+12x + 11}{(x + 3)^{2}}$
(e) $\frac{3x^{2}+12x + 11}{(x + 3)^{2}}$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x^{2}+3x + 2$, so $u'=2x + 3$, and $v=x + 3$, so $v'=1$.

Step2: Substitute into quotient - rule formula

$f'(x)=\frac{(2x + 3)(x + 3)-(x^{2}+3x + 2)\times1}{(x + 3)^{2}}$.

Step3: Expand the numerator

First, expand $(2x + 3)(x + 3)=2x^{2}+6x+3x + 9=2x^{2}+9x + 9$. Then, $(x^{2}+3x + 2)\times1=x^{2}+3x + 2$. So the numerator is $2x^{2}+9x + 9-(x^{2}+3x + 2)$.

Step4: Simplify the numerator

$2x^{2}+9x + 9-(x^{2}+3x + 2)=2x^{2}+9x + 9 - x^{2}-3x - 2=x^{2}+6x + 7$.

Answer:

C. $\frac{x^{2}+6x + 7}{(x + 3)^{2}}$