$$\text { Find the value of } ( f \circ g ) ^ { \prime } \text { at the given value of } x \text {. }$$
-$$f ( u ) = \frac { u - 1 } { u + 1 } , u = g ( x ) = \sqrt { x } , x = 49$$
A) $\frac { 1 } { 49 }$
B) $\frac { 1 } { 7 }$
C) $\frac { 1 } { 448 }$
D) $\frac { 1 } { 64 }$

Question

Quizplus · Accepted Answer

To find $(f \circ g)'(x)$ at $x = 49$, we first need to find the derivative of the composite function $f(g(x))$ and then substitute $x = 49$. Given $f(u) = \frac{u - 1}{u + 1}$ and $u = g(x) = \sqrt{x}$, we have $f(g(x)) = \frac{\sqrt{x} - 1}{\sqrt{x} + 1}$.The derivative of $f(g(x))$ with respect to $x$ can be found using the chain rule: $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$.First, find $f'(u)$:$$f'(u) = \frac{d}{du}\left(\frac{u - 1}{u + 1}\right) = \frac{(u + 1)(1) - (u - 1)(1)}{(u + 1)^2} = \frac{2}{(u + 1)^2}$$Then, find $g'(x)$:$$g'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$$Substitute $u = g(x) = \sqrt{x}$ into $f'(u)$ and multiply by $g'(x)$:$$(f \circ g)'(x) = \frac{2}{(\sqrt{x} + 1)^2} \cdot \frac{1}{2\sqrt{x}}$$Substitute $x = 49$:$$(f \circ g)'(49) = \frac{2}{(\sqrt{49} + 1)^2} \cdot \frac{1}{2\sqrt{49}} = \frac{2}{(7 + 1)^2} \cdot \frac{1}{2 \cdot 7} = \frac{2}{64} \cdot \frac{1}{14} = \frac{1}{32} \cdot \frac{1}{14} = \frac{1}{448}$$Therefore, the correct answer is $\frac{1}{448}$, which is option C.

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.

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