Question

find $\frac{dy}{dx}$ b. $y = \frac{x^{2}+1}{x - 3}$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = x^{2}+1$, so $u'=2x$, and $v=x - 3$, so $v' = 1$.

Step2: Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule formula

$\frac{dy}{dx}=\frac{(2x)(x - 3)-(x^{2}+1)\times1}{(x - 3)^{2}}$.

Step3: Expand the numerator

Expand $(2x)(x - 3)-(x^{2}+1)$:
$(2x)(x - 3)=2x^{2}-6x$, so $(2x)(x - 3)-(x^{2}+1)=2x^{2}-6x - x^{2}-1$.
Combining like - terms in the numerator gives $x^{2}-6x - 1$.

Answer:

$\frac{x^{2}-6x - 1}{(x - 3)^{2}}$