Question

consider the following.

$f(x)=\frac{x}{x - 3}$

find the first derivative of the function.

$f(x)=square$

find the second derivative of the function.

$f(x)=square$

Explanation:

Step1: Apply quotient - rule for first derivative

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x$, $u'=1$, $v=x - 3$, and $v' = 1$. So, $f'(x)=\frac{1\times(x - 3)-x\times1}{(x - 3)^{2}}$.

Step2: Simplify the first - derivative expression

$f'(x)=\frac{x-3 - x}{(x - 3)^{2}}=\frac{-3}{(x - 3)^{2}}=-3(x - 3)^{-2}$.

Step3: Apply chain - rule for second derivative

The chain - rule states that if $y = u^{n}$, then $y'=nu^{n - 1}u'$. Here, $u=(x - 3)$, $n=-2$, and $u' = 1$. So, $f''(x)=(-3)\times(-2)(x - 3)^{-3}\times1$.

Step4: Simplify the second - derivative expression

$f''(x)=\frac{6}{(x - 3)^{3}}$.

Answer:

$f'(x)=-\frac{3}{(x - 3)^{2}}$
$f''(x)=\frac{6}{(x - 3)^{3}}$