find the derivative of the function.
$y = \frac{4x^{2}+3}{x^{2}+1}$
$y=\square$

Question

Accepted Answer

# Explanation:
## Step1: Apply the quotient rule
The quotient rule states that if $y = \frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4x^{2}+3$, so $u'=8x$; $v=x^{2}+1$, so $v' = 2x$.
## Step2: Substitute $u$, $u'$, $v$, $v'$ into the quotient rule formula
$$
\begin{align*}
y'&=\frac{(8x)(x^{2}+1)-(4x^{2}+3)(2x)}{(x^{2}+1)^{2}}\
&=\frac{8x^{3}+8x-(8x^{3}+6x)}{(x^{2}+1)^{2}}\
&=\frac{8x^{3}+8x - 8x^{3}-6x}{(x^{2}+1)^{2}}\
&=\frac{2x}{(x^{2}+1)^{2}}
\end{align*}
$$

# Answer:
$\frac{2x}{(x^{2}+1)^{2}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the quotient rule

Step2: Substitute \(u\), \(u'\), \(v\), \(v'\) into the quotient rule formula

Answer: