find the derivative.\n$\frac{(2x - 2)(2x + 4)}{x + 4}$\n\na. $\frac{12x^{2}+40x + 8}{(x + 4)^{2}}$\nb…

find the derivative.\n$\frac{(2x - 2)(2x + 4)}{x + 4}$\n\na. $\frac{12x^{2}+40x + 8}{(x + 4)^{2}}$\nb. $\frac{4x^{2}+32x + 24}{x + 4}$\nc. $\frac{4x^{2}+32x + 24}{(x + 4)^{2}}$\nd. $\frac{12x^{2}+40x + 8}{x + 4}$

find the derivative.\n$\frac{(2x - 2)(2x + 4)}{x + 4}$\n\na. $\frac{12x^{2}+40x + 8}{(x + 4)^{2}}$\nb. $\frac{4x^{2}+32x + 24}{x + 4}$\nc. $\frac{4x^{2}+32x + 24}{(x + 4)^{2}}$\nd. $\frac{12x^{2}+40x + 8}{x + 4}$

Answer

Explanation:

Step1: Expand the numerator

First, expand ((2x - 2)(2x+4)) using FOIL method. ((2x - 2)(2x + 4)=4x^{2}+8x-4x - 8=4x^{2}+4x - 8). So the function (y=\frac{4x^{2}+4x - 8}{x + 4}).

Step2: Apply the quotient - rule

The quotient - rule states that if (y=\frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 4x^{2}+4x - 8), (u^\prime=8x + 4), (v=x + 4), and (v^\prime=1). [ \begin{align*} y^\prime&=\frac{(8x + 4)(x + 4)-(4x^{2}+4x - 8)\times1}{(x + 4)^{2}}\ &=\frac{8x^{2}+32x+4x + 16-(4x^{2}+4x - 8)}{(x + 4)^{2}}\ &=\frac{8x^{2}+36x + 16-4x^{2}-4x + 8}{(x + 4)^{2}}\ &=\frac{(8x^{2}-4x^{2})+(36x-4x)+(16 + 8)}{(x + 4)^{2}}\ &=\frac{4x^{2}+32x + 24}{(x + 4)^{2}} \end{align*} ]

Answer:

C. (\frac{4x^{2}+32x + 24}{(x + 4)^{2}})