Find the derivative of $g(x)=\frac{x}{f(x)}$.
Select one:
a. $g^{\prime}(x)=\frac{1-f(x)}{[f(x)]^{2}}$
b. $g^{\prime}(x)=\frac{x-f(x)}{[f(x)]^{2}}$
c. $g^{\prime}(x)=f^{\prime}(x)$
d. $g^{\prime}(x)=\frac{f(x)-x f^{\prime}(x)}{[f(x)]^{2}}$
$g^{\prime}(x)=\frac{f^{\prime}(x)}{[f(x)]^{2}}$

Answer

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Answer

$\boxed{g^\prime(x)=\frac{f(x)-x f^\prime(x)}{[f(x)]^{2}}}$ is the final answer.

Steps

Step 1 ：Let $u=x$ and $v=f(x)$.

Step 2 ：The derivative of $u=x$ is $u'=1$.

Step 3 ：The derivative of $v=f(x)$ is $v'=f'(x)$.

Step 4 ：Substitute these into the quotient rule to get $g'(x)=\frac{f(x)(1)-x f'(x)}{[f(x)]^{2}}$.

Step 5 ：$\boxed{g^\prime(x)=\frac{f(x)-x f^\prime(x)}{[f(x)]^{2}}}$ is the final answer.