differentiate. g(x)=\frac{8x - 1}{6x + 7}+x^{3} g(x)=\frac{62}{(6x + 7)^{2}}+3x^{2}

Explanation:

Step1: Apply quotient - rule to $\frac{8x - 1}{6x+7}$

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 8x-1$, $u^\prime=8$, $v = 6x + 7$, $v^\prime=6$. So, $\left(\frac{8x - 1}{6x+7}\right)^\prime=\frac{8(6x + 7)-(8x - 1)\times6}{(6x + 7)^{2}}=\frac{48x+56-(48x - 6)}{(6x + 7)^{2}}=\frac{48x+56 - 48x + 6}{(6x + 7)^{2}}=\frac{62}{(6x + 7)^{2}}$.

Step2: Differentiate $x^{3}$

Using the power - rule $\left(x^{n}\right)^\prime=nx^{n - 1}$, for $n = 3$, $\left(x^{3}\right)^\prime=3x^{2}$.

Step3: Combine the derivatives

Since $g(x)=\frac{8x - 1}{6x+7}+x^{3}$, by the sum - rule of differentiation $(f(x)+h(x))^\prime=f^\prime(x)+h^\prime(x)$, $g^\prime(x)=\left(\frac{8x - 1}{6x+7}\right)^\prime+\left(x^{3}\right)^\prime=\frac{62}{(6x + 7)^{2}}+3x^{2}$.

Answer:

$g^\prime(x)=\frac{62}{(6x + 7)^{2}}+3x^{2}$