PART 10：例題-反函數的微分法一

設 \(f(x) = \frac{x}{{x + 1}}\;,x > 0\) 為一對一函數，求 \({\left( {{f^{ - 1}}} \right)^\prime }\left( {1/3} \right)\)

SOL:

找出 \({f^{ - 1}}\left( x \right)\) 再微分代入 \(1/3\) 求值

(1) \(y = \frac{x}{{x + 1}}\;\)

(2) 解 \(x\) ， \(y = \frac{x}{{x + 1}}\;\)  \(\Rightarrow yx + y = x\;\; \Rightarrow \;x = \frac{y}{{1 - y}}\)

(3)  \(x\) 與 \(y\) 互換， \(\;y = \frac{x}{{1 - x}}\) (這就是反函數 \({f^{ - 1}}\left( x \right)\))

(4)  \({f^{ - 1}}\left( x \right) = \frac{x}{{1 - x}}\) ， \({\left( {{f^{ - 1}}} \right)^\prime }\left( x \right) = \frac{{(1 - x) - x( - 1)}}{{{{(1 - x)}^2}}} = \frac{1}{{{{(1 - x)}^2}}}\)

(5) \({\left( {{f^{ - 1}}} \right)^\prime }\left( {1/3} \right) = \frac{1}{{{{(1 - 1/3)}^2}}}\) = \(9/4\)