Question

hw 12 - higher order derivatives section 2 (1 point)
let $f(x)=\frac{9x}{2 - 5x}$. then
$f(x)=square$
and
$f(x)=square$
note: you can earn partial credit on this problem.
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Explanation:

Step1: Apply quotient - rule for first derivative

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = 9x$, $u^{\prime}(x)=9$, $v(x)=2 - 5x$, and $v^{\prime}(x)=-5$.
\[

$$\begin{align*} f^{\prime}(x)&=\frac{9(2 - 5x)-9x(-5)}{(2 - 5x)^{2}}\\ &=\frac{18-45x + 45x}{(2 - 5x)^{2}}\\ &=\frac{18}{(2 - 5x)^{2}} \end{align*}$$

\]

Step2: Apply quotient - rule for second derivative

Now, consider $f^{\prime}(x)=\frac{18}{(2 - 5x)^{2}}=\frac{18}{v^{2}(x)}$ where $v(x)=2 - 5x$ and $v^{\prime}(x)=-5$. Using the quotient - rule $f^{\prime\prime}(x)=\frac{0\times(2 - 5x)^{2}-18\times2(2 - 5x)(-5)}{(2 - 5x)^{4}}$.
\[

$$\begin{align*} f^{\prime\prime}(x)&=\frac{180(2 - 5x)}{(2 - 5x)^{4}}\\ &=\frac{180}{(2 - 5x)^{3}} \end{align*}$$

\]

Answer:

$f^{\prime}(x)=\frac{18}{(2 - 5x)^{2}}$
$f^{\prime\prime}(x)=\frac{180}{(2 - 5x)^{3}}$