Question

section 2.5: product and quotient rules (h
score: 20/80 answered: 2/8
question 3
if $f(x)=\frac{6 - x^{2}}{6 + x^{2}}$, find:
$f(x)=$
question help: video

Explanation:

Step1: Identify quotient - rule

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

Step2: Find $u'$ and $v'$

Differentiate $u$ with respect to $x$: $u'=\frac{d}{dx}(6 - x^{2})=-2x$. Differentiate $v$ with respect to $x$: $v'=\frac{d}{dx}(6 + x^{2}) = 2x$.

Step3: Apply the quotient - rule

$f'(x)=\frac{u'v - uv'}{v^{2}}=\frac{(-2x)(6 + x^{2})-(6 - x^{2})(2x)}{(6 + x^{2})^{2}}$.

Step4: Expand and simplify

\[

$$\begin{align*} f'(x)&=\frac{-12x-2x^{3}-12x + 2x^{3}}{(6 + x^{2})^{2}}\\ &=\frac{-24x}{(6 + x^{2})^{2}} \end{align*}$$

\]

Answer:

$\frac{-24x}{(6 + x^{2})^{2}}$