Question

find the derivative of the function.
y = (8x^2 + 1)/(x^2 + 2)
y = □

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 8x^{2}+1$, $u'=16x$, $v=x^{2}+2$, and $v' = 2x$.

Step2: Substitute into quotient - rule

$y'=\frac{(16x)(x^{2}+2)-(8x^{2}+1)(2x)}{(x^{2}+2)^{2}}$.

Step3: Expand the numerator

Expand $(16x)(x^{2}+2)=16x^{3}+32x$ and $(8x^{2}+1)(2x)=16x^{3}+2x$. Then the numerator is $(16x^{3}+32x)-(16x^{3}+2x)$.

Step4: Simplify the numerator

$(16x^{3}+32x)-(16x^{3}+2x)=16x^{3}+32x - 16x^{3}-2x=30x$.

Answer:

$\frac{30x}{(x^{2}+2)^{2}}$