Question

question 2. find the derivatives of the following functions using the quotient rule.
(a) $r(x)=\frac{3x^{2}+5x}{x + 4}$
(b) $s(t)=\frac{3t^{4}+2}{t^{3}+1}$

Explanation:

Step1: Recall quotient - rule formula

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

Step2: Identify $u$, $v$, $u'$ and $v'$ for $r(x)=\frac{3x^{2}+5x}{x + 4}$

Let $u = 3x^{2}+5x$, then $u'=6x + 5$; let $v=x + 4$, then $v'=1$.

Step3: Apply the quotient - rule

\[

$$\begin{align*} r'(x)&=\frac{(6x + 5)(x + 4)-(3x^{2}+5x)\times1}{(x + 4)^{2}}\\ &=\frac{6x^{2}+24x+5x + 20-(3x^{2}+5x)}{(x + 4)^{2}}\\ &=\frac{6x^{2}+29x + 20-3x^{2}-5x}{(x + 4)^{2}}\\ &=\frac{3x^{2}+24x + 20}{(x + 4)^{2}} \end{align*}$$

\]

Step4: Identify $u$, $v$, $u'$ and $v'$ for $s(t)=\frac{3t^{4}+2}{t^{3}+1}$

Let $u = 3t^{4}+2$, then $u'=12t^{3}$; let $v=t^{3}+1$, then $v'=3t^{2}$.

Step5: Apply the quotient - rule

\[

$$\begin{align*} s'(t)&=\frac{12t^{3}(t^{3}+1)-(3t^{4}+2)\times3t^{2}}{(t^{3}+1)^{2}}\\ &=\frac{12t^{6}+12t^{3}-(9t^{6}+6t^{2})}{(t^{3}+1)^{2}}\\ &=\frac{12t^{6}+12t^{3}-9t^{6}-6t^{2}}{(t^{3}+1)^{2}}\\ &=\frac{3t^{6}+12t^{3}-6t^{2}}{(t^{3}+1)^{2}} \end{align*}$$

\]

Answer:

(a) $r'(x)=\frac{3x^{2}+24x + 20}{(x + 4)^{2}}$
(b) $s'(t)=\frac{3t^{6}+12t^{3}-6t^{2}}{(t^{3}+1)^{2}}$