Question

72 - 73. first and second derivatives find $f(x)$ and $f(x)$. 72. $f(x)=\frac{x}{x + 2}$ 73. $f(x)=\frac{x^{2}-7x}{x + 1}$

Explanation:

Step1: Use quotient - rule for first derivative

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. For $f(x)=\frac{x}{x + 2}$, where $u = x$, $u'=1$, $v=x + 2$, $v'=1$.
\[

$$\begin{align*} f'(x)&=\frac{1\cdot(x + 2)-x\cdot1}{(x + 2)^{2}}\\ &=\frac{x+2 - x}{(x + 2)^{2}}\\ &=\frac{2}{(x + 2)^{2}} \end{align*}$$

\]

Step2: Use quotient - rule for second derivative

Now, for $y = f'(x)=\frac{2}{(x + 2)^{2}}=\frac{2}{(x + 2)^{2}}$, where $u = 2$, $u'=0$, $v=(x + 2)^{2}$, $v'=2(x + 2)$.
\[

$$\begin{align*} f''(x)&=\frac{0\cdot(x + 2)^{2}-2\cdot2(x + 2)}{(x + 2)^{4}}\\ &=\frac{-4(x + 2)}{(x + 2)^{4}}\\ &=-\frac{4}{(x + 2)^{3}} \end{align*}$$

\]

Answer:

For $f(x)=\frac{x}{x + 2}$, $f'(x)=\frac{2}{(x + 2)^{2}}$, $f''(x)=-\frac{4}{(x + 2)^{3}}$

Now for $f(x)=\frac{x^{2}-7x}{x + 1}$:

Step1: Use quotient - rule for first derivative

Let $u=x^{2}-7x$, $u'=2x-7$, $v=x + 1$, $v'=1$.
\[

$$\begin{align*} f'(x)&=\frac{(2x - 7)(x + 1)-(x^{2}-7x)\cdot1}{(x + 1)^{2}}\\ &=\frac{2x^{2}+2x-7x-7-(x^{2}-7x)}{(x + 1)^{2}}\\ &=\frac{2x^{2}+2x-7x-7 - x^{2}+7x}{(x + 1)^{2}}\\ &=\frac{x^{2}+2x-7}{(x + 1)^{2}} \end{align*}$$

\]

Step2: Use quotient - rule for second derivative

Let $u=x^{2}+2x - 7$, $u'=2x+2$, $v=(x + 1)^{2}$, $v'=2(x + 1)$.
\[

$$\begin{align*} f''(x)&=\frac{(2x + 2)(x + 1)^{2}-(x^{2}+2x - 7)\cdot2(x + 1)}{(x + 1)^{4}}\\ &=\frac{(x + 1)[(2x + 2)(x + 1)-2(x^{2}+2x - 7)]}{(x + 1)^{4}}\\ &=\frac{(2x^{2}+2x+2x + 2)-(2x^{2}+4x-14)}{(x + 1)^{3}}\\ &=\frac{2x^{2}+4x + 2-2x^{2}-4x + 14}{(x + 1)^{3}}\\ &=\frac{16}{(x + 1)^{3}} \end{align*}$$

\]