Question

1. let (f(u)=\frac{u + 1}{u - 1}) and (g(x)=u=sqrt{x}). find ((fcirc g)(4)).

Explanation:

Step1: Find the composition $f\circ g(x)$

First, substitute $u = g(x)=\sqrt{x}$ into $f(u)$. So, $(f\circ g)(x)=f(g(x))=\frac{\sqrt{x}+1}{\sqrt{x}-1}$.

Step2: Differentiate $(f\circ g)(x)$ using the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Let $u=\sqrt{x}+1=x^{\frac{1}{2}}+1$ and $v=\sqrt{x}-1=x^{\frac{1}{2}}-1$. Then $u'=\frac{1}{2}x^{-\frac{1}{2}}$ and $v'=\frac{1}{2}x^{-\frac{1}{2}}$.
\[

$$\begin{align*} (f\circ g)'(x)&=\frac{\frac{1}{2}x^{-\frac{1}{2}}(\sqrt{x}-1)-\frac{1}{2}x^{-\frac{1}{2}}(\sqrt{x}+1)}{(\sqrt{x}-1)^{2}}\\ &=\frac{\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}-\frac{1}{2}x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{1}{2}}\sqrt{x}-\frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x}-1)^{2}}\\ &=\frac{-x^{-\frac{1}{2}}}{(\sqrt{x}-1)^{2}}=\frac{-1}{\sqrt{x}(\sqrt{x}-1)^{2}} \end{align*}$$

\]

Step3: Evaluate $(f\circ g)'(x)$ at $x = 4$

Substitute $x = 4$ into $(f\circ g)'(x)$. When $x = 4$, $\sqrt{x}=2$.
\[

$$\begin{align*} (f\circ g)'(4)&=\frac{-1}{\sqrt{4}(2 - 1)^{2}}\\ &=\frac{-1}{2\times1^{2}}=-\frac{1}{2} \end{align*}$$

\]

Answer:

$-\frac{1}{2}$