Re: [微積] 反函數
※ 引述《dream0402 (5656565656)》之銘言：
: 已知 f[g(x)]=x 且 f'(x)=1+f^2(x) 則g'(x)=?
: 先感謝各位幫忙解答~~~
g' * df/dg = 1
g' * [1 + (f(g))^2] = 1
g'(x) = 1 / [1 + x^2]
g(x) = arctan(x) + C
f(x) = tan(g(x) - C)
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