Question

consider the following.
f(u) = 7u^4 - 2u
h(u) = u + 1
find f(u) and h(u).
f(u) = 28u^3 - 2
well done!
h(u) = 1
good work!
differentiate.
g(u)=\frac{7u^4 - 2u}{u + 1}
g(u)=\frac{-21u^4 - 28u^3 + 2}{(7u^4 - 2u)^2}
resources

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $G(u)=\frac{f(u)}{h(u)}$, then $G^{\prime}(u)=\frac{f^{\prime}(u)h(u)-f(u)h^{\prime}(u)}{(h(u))^{2}}$. Here, $f(u) = 7u^{4}-2u$, $f^{\prime}(u)=28u^{3}-2$, $h(u)=u + 1$, and $h^{\prime}(u)=1$.

Step2: Substitute functions and their derivatives

Substitute the values of $f(u)$, $f^{\prime}(u)$, $h(u)$, and $h^{\prime}(u)$ into the quotient - rule formula.
$G^{\prime}(u)=\frac{(28u^{3}-2)(u + 1)-(7u^{4}-2u)\times1}{(u + 1)^{2}}$.

Step3: Expand the numerator

Expand $(28u^{3}-2)(u + 1)$ and simplify the numerator.
$(28u^{3}-2)(u + 1)=28u^{4}+28u^{3}-2u - 2$.
The numerator becomes $28u^{4}+28u^{3}-2u - 2-(7u^{4}-2u)=28u^{4}+28u^{3}-2u - 2 - 7u^{4}+2u=21u^{4}+28u^{3}-2$.
So, $G^{\prime}(u)=\frac{21u^{4}+28u^{3}-2}{(u + 1)^{2}}$.

Answer:

$G^{\prime}(u)=\frac{21u^{4}+28u^{3}-2}{(u + 1)^{2}}$