Question

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

Explanation:

Step1: Apply quotient - rule to first term

Let $u = 2x - 1$, $v=7x + 8$. Then $\frac{d}{dx}(\frac{u}{v})=\frac{u'v - uv'}{v^{2}}$, where $u' = 2$ and $v'=7$. So $\frac{d}{dx}(\frac{2x - 1}{7x + 8})=\frac{2(7x + 8)-7(2x - 1)}{(7x + 8)^{2}}=\frac{14x+16 - 14x + 7}{(7x + 8)^{2}}=\frac{23}{(7x + 8)^{2}}$.

Step2: Differentiate second term

$\frac{d}{dx}(x^{3}) = 3x^{2}$.

Step3: Sum the derivatives

$g'(x)=\frac{23}{(7x + 8)^{2}}+3x^{2}$.

Answer:

$\frac{23}{(7x + 8)^{2}}+3x^{2}$