Question

differentiate the function.
y = \frac{4x^{2}-3}{2x^{3}+7}
y = square

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4x^{2}-3$, $u'=8x$, $v = 2x^{3}+7$, $v'=6x^{2}$.

Step2: Substitute into formula

$y'=\frac{(8x)(2x^{3}+7)-(4x^{2}-3)(6x^{2})}{(2x^{3}+7)^{2}}=\frac{16x^{4}+56x - 24x^{4}+18x^{2}}{(2x^{3}+7)^{2}}=\frac{- 8x^{4}+18x^{2}+56x}{(2x^{3}+7)^{2}}$

Answer:

$\frac{- 8x^{4}+18x^{2}+56x}{(2x^{3}+7)^{2}}$