let (f(x)=\frac{x}{x - 1}). calculate (g(x)), where (g(x)) is the inverse of (f(x)). (g(x)=)

Explanation:

Step1: Recall inverse - function derivative formula

If $y = f(x)$ has an inverse function $x = g(y)$ (where $g$ is the inverse of $f$), then $g^{\prime}(y)=\frac{1}{f^{\prime}(g(y))}$. First, we need to find $f^{\prime}(x)$.

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

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. For $f(x)=\frac{x}{x - 1}$, where $u(x)=x$, $u^{\prime}(x)=1$, $v(x)=x - 1$, and $v^{\prime}(x)=1$. Then $f^{\prime}(x)=\frac{1\cdot(x - 1)-x\cdot1}{(x - 1)^{2}}=\frac{x - 1-x}{(x - 1)^{2}}=-\frac{1}{(x - 1)^{2}}$.

Step3: Use the inverse - function derivative formula

Let $y = f(x)$. Then $g^{\prime}(y)=\frac{1}{f^{\prime}(g(y))}$. Since $y=\frac{x}{x - 1}$, we can solve for $x$ in terms of $y$: [ \begin{align*} y&=\frac{x}{x - 1}\ y(x - 1)&=x\ yx-y&=x\ yx-x&=y\ x(y - 1)&=y\ x = g(y)&=\frac{y}{y - 1} \end{align*} ] Substitute $x = g(y)$ into $f^{\prime}(x)$: $f^{\prime}(g(y))=-\frac{1}{(\frac{y}{y - 1}-1)^{2}}$. [ \begin{align*} \frac{y}{y - 1}-1&=\frac{y-(y - 1)}{y - 1}=\frac{y - y + 1}{y - 1}=\frac{1}{y - 1}\ f^{\prime}(g(y))&=-(y - 1)^{2} \end{align*} ] So $g^{\prime}(y)=(y - 1)^{2}$. Replacing $y$ with $x$, we get $g^{\prime}(x)=(x - 1)^{2}$.

Answer:

$(x - 1)^{2}$