Question

differentiate the function.
y = \frac{3x^{2}-5}{5x^{3}+2}
y = \square

Explanation:

Step1: Apply quotient - rule

Quotient rule: $(\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}$, where $u = 3x^{2}-5$, $u'=6x$, $v = 5x^{3}+2$, $v' = 15x^{2}$.

Step2: Substitute values

$y'=\frac{(6x)(5x^{3}+2)-(3x^{2}-5)(15x^{2})}{(5x^{3}+2)^{2}}$.

Step3: Expand expressions

$y'=\frac{30x^{4}+12x-(45x^{4}-75x^{2})}{(5x^{3}+2)^{2}}=\frac{30x^{4}+12x - 45x^{4}+75x^{2}}{(5x^{3}+2)^{2}}$.

Step4: Simplify

$y'=\frac{- 15x^{4}+75x^{2}+12x}{(5x^{3}+2)^{2}}$.

Answer:

$\frac{-15x^{4}+75x^{2}+12x}{(5x^{3}+2)^{2}}$