Problem
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. $g^{\prime}(x)=\frac{f(x)-x f^{\prime}(x)}{[f(x)]^{2}}$ $g^{\prime}(x)=\frac{f^{\prime}(x)}{[f(x)]^{2}}$
Solution
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.
