Question

homework 3.3 differentiation rules
score: 80/120 answered: 8/12
question 9
textbook videos •
if (f(x)=\frac{7 - x^{2}}{8 + x^{3}}), find:
f(x) =
question help: video

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 7 - x^{2}$, so $u^\prime=-2x$, and $v = 8 + x^{3}$, so $v^\prime = 3x^{2}$.

Step2: Apply the quotient - rule

$f^\prime(x)=\frac{(-2x)(8 + x^{3})-(7 - x^{2})(3x^{2})}{(8 + x^{3})^{2}}$.

Step3: Expand the numerator

First, expand $(-2x)(8 + x^{3})=-16x-2x^{4}$. Second, expand $(7 - x^{2})(3x^{2}) = 21x^{2}-3x^{4}$. Then the numerator is $-16x-2x^{4}-(21x^{2}-3x^{4})=-16x-2x^{4}-21x^{2}+3x^{4}=x^{4}-21x^{2}-16x$.

Answer:

$\frac{x^{4}-21x^{2}-16x}{(8 + x^{3})^{2}}$