hw11 differentiation rules ii (target c1, c2, c5; §3.3)\nscore: 4/8 answered: 4/8\nquestion 5\nif…

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 6 - x^{2}$, $u'=-2x$, $v = 7 + x^{2}$, and $v' = 2x$.

Step2: Apply the quotient - rule

$f'(x)=\frac{(-2x)(7 + x^{2})-(6 - x^{2})(2x)}{(7 + x^{2})^{2}}$.

Step3: Expand the numerator

First, expand $(-2x)(7 + x^{2})=-14x-2x^{3}$ and $(6 - x^{2})(2x)=12x - 2x^{3}$. Then $(-2x)(7 + x^{2})-(6 - x^{2})(2x)=(-14x-2x^{3})-(12x - 2x^{3})$.

Step4: Simplify the numerator

$(-14x-2x^{3})-(12x - 2x^{3})=-14x-2x^{3}-12x + 2x^{3}=-26x$.

Answer:

$\frac{-26x}{(7 + x^{2})^{2}}$