Question

question if $q(x)=ln(6x^{2}+2x + 2)$, find $q(x)$. select the correct answer below: $\frac{12x + 2}{ln(6x^{2}+2x+2)}$ $ln(12x + 2)$ $\frac{1}{6x^{2}+2x+2}$ $\frac{3x^{2}+x + 1}{6x+1}$ $\frac{6x+1}{3x^{2}+x+1}$

Explanation:

Step1: Recall chain - rule

The chain - rule states that if $y = \ln(u)$ and $u = g(x)$, then $y^\prime=\frac{u^\prime}{u}$. Here, $u = 6x^{2}+2x + 2$ and $y = q(x)=\ln(u)$.

Step2: Differentiate $u$

Differentiate $u = 6x^{2}+2x + 2$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u^\prime=\frac{d}{dx}(6x^{2}+2x + 2)=12x + 2$.

Step3: Apply chain - rule

Since $q(x)=\ln(u)$ and $u = 6x^{2}+2x + 2$, by the chain - rule $q^\prime(x)=\frac{u^\prime}{u}=\frac{12x + 2}{6x^{2}+2x + 2}$.

Answer:

The correct answer is $\frac{12x + 2}{6x^{2}+2x + 2}$ (none of the given options are correct based on the above - derived result). If we simplify $\frac{12x + 2}{6x^{2}+2x + 2}=\frac{2(6x + 1)}{2(3x^{2}+x + 1)}=\frac{6x+1}{3x^{2}+x + 1}$, so the answer is $\boxed{\text{O }\frac{6x + 1}{3x^{2}+x + 1}}$.