$$\text { Find the value of } ( f \circ g ) ^ { \prime } \text { at the given value of } x \text {. }$$
-$$f ( u ) = u ^ { 3 } , u = g ( x ) = \frac { x + 4 } { x - 2 } , x = 0$$
A) 18
B) - 18
C) 12
D) - 12

Question

Quizplus · Accepted Answer

To find the derivative of the composite function $(f \circ g)(x)$ at $x = 0$, we first need to apply the chain rule. The chain rule states that $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$.Given $f(u) = u^3$ and $u = g(x) = \frac{x + 4}{x - 2}$, we first find the derivative of each function.1. The derivative of $f(u) = u^3$ with respect to $u$ is $f'(u) = 3u^2$.2. To find $g'(x)$, we differentiate $g(x) = \frac{x + 4}{x - 2}$ with respect to $x$. Using the quotient rule, $g'(x) = \frac{(1)(x - 2) - (x + 4)(1)}{(x - 2)^2} = \frac{-6}{(x - 2)^2}$.Now, we apply the chain rule:$(f \circ g)'(x) = f'(g(x)) \cdot g'(x) = 3(g(x))^2 \cdot g'(x)$.Substituting $g(x) = \frac{x + 4}{x - 2}$ and $g'(x) = \frac{-6}{(x - 2)^2}$ into the equation, we get:$(f \circ g)'(x) = 3\left(\frac{x + 4}{x - 2}\right)^2 \cdot \frac{-6}{(x - 2)^2}$.Finally, substitute $x = 0$ into the equation:$(f \circ g)'(0) = 3\left(\frac{0 + 4}{0 - 2}\right)^2 \cdot \frac{-6}{(0 - 2)^2} = 3\left(\frac{4}{-2}\right)^2 \cdot \frac{-6}{4} = 3 \cdot 4 \cdot (-\frac{3}{2}) = -18$.Therefore, the correct answer is $-18$, which is choice B.

Find the value of (f∘g)′ at the given value of x. \text { Find the value of } ( f \circ g ) ^ { \prime } \text { at the given value of } x \text {. } Find the value of (f∘g)′ at the given value of x.

$\text { Find the value of } ( f \circ g ) ^ { \prime } \text { at the given value of } x \text {. }$