Question

find h(x) where f(x) is an unspecified differentiable function.
h(x)=\frac{ln x^{7}}{f(x)}

choose the correct answer below.

a. h(x)=\frac{ln x^{7}(f(x)-f(x))}{f(x)^{2}}
b. h(x)=\frac{7f(x)-(xln x^{7})f(x)}{xf(x)^{2}}
c. h(x)=\frac{ln x^{7}(f(x)-f(x))}{(ln x^{7})^{2}}
d. h(x)=\frac{(xln x^{7})f(x)-7f(x)}{x(ln x^{7})^{2}}

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x)=\ln x^{7}$ and $v(x)=f(x)$.

Step2: Differentiate $u(x)$

First, simplify $u(x)=\ln x^{7}=7\ln x$. Then, using the rule $\frac{d}{dx}(\ln x)=\frac{1}{x}$, we get $u^{\prime}(x)=\frac{7}{x}$. And $v^{\prime}(x)=f^{\prime}(x)$.

Step3: Apply quotient - rule

Substitute $u(x), u^{\prime}(x), v(x), v^{\prime}(x)$ into the quotient - rule formula:
\[

$$\begin{align*} h^{\prime}(x)&=\frac{\frac{7}{x}\cdot f(x)-7\ln x\cdot f^{\prime}(x)}{f(x)^2}\\ &=\frac{7f(x)-(x\ln x^{7})f^{\prime}(x)}{x f(x)^2} \end{align*}$$

\]

Answer:

B. $h^{\prime}(x)=\frac{7f(x)-(x\ln x^{7})f^{\prime}(x)}{x f(x)^2}$