Question

let (f) be a differentiable function such that (f(8)=2) and (f^{prime}(8)=5). if (g) is the function defined by (g(x)=\frac{f(x)}{sqrt3{x}}), what is the value of (g^{prime}(8))?
(a) (\frac{59}{24})
(b) (\frac{5}{2})
(c) (\frac{61}{24})
(d) 60

Explanation:

Step1: Apply quotient - rule for differentiation

The quotient - rule states that if $g(x)=\frac{u(x)}{v(x)}$, then $g^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = f(x)$ and $v(x)=x^{\frac{1}{3}}$. So $u^{\prime}(x)=f^{\prime}(x)$ and $v^{\prime}(x)=\frac{1}{3}x^{-\frac{2}{3}}$. Then $g^{\prime}(x)=\frac{f^{\prime}(x)x^{\frac{1}{3}}-f(x)\cdot\frac{1}{3}x^{-\frac{2}{3}}}{(x^{\frac{1}{3}})^2}=\frac{f^{\prime}(x)x^{\frac{1}{3}}-\frac{1}{3}f(x)x^{-\frac{2}{3}}}{x^{\frac{2}{3}}}$.

Step2: Substitute $x = 8$

We know that $f(8) = 2$, $f^{\prime}(8)=5$, and $x = 8$. First, calculate the values of $x^{\frac{1}{3}}$ and $x^{-\frac{2}{3}}$ when $x = 8$. Since $8^{\frac{1}{3}}=2$ and $8^{-\frac{2}{3}}=\frac{1}{4}$. Substitute these values into the formula for $g^{\prime}(x)$:
\[

$$\begin{align*} g^{\prime}(8)&=\frac{f^{\prime}(8)\cdot8^{\frac{1}{3}}-\frac{1}{3}f(8)\cdot8^{-\frac{2}{3}}}{8^{\frac{2}{3}}}\\ &=\frac{5\times2-\frac{1}{3}\times2\times\frac{1}{4}}{4}\\ &=\frac{10-\frac{1}{6}}{4}\\ &=\frac{\frac{60 - 1}{6}}{4}\\ &=\frac{\frac{59}{6}}{4}\\ &=\frac{59}{24} \end{align*}$$

\]

Answer:

A. $\frac{59}{24}$