題目 2. (10 points) Define f(x)=tan2(x)f(x)=\tan^2(x)f(x)=tan2(x) for x∈(0,π2)x\in\left(0,\dfrac{\pi}{2}\right)x∈(0,2π) and let f−1f^{-1}f−1 be its inverse function. Find (f−1)′(3)(f^{-1})'(3)(f−1)′(3). 解答 解法一 若 f(x)=3f(x)=3f(x)=3，則 tanx=3\tan x=\sqrt3tanx=3；由 x∈(0,π/2)x\in(0,\pi/2)x∈(0,π/2) 得 x=π/3x=\pi/3x=π/3。反函數導數公式給出 (f−1)′(3)=1f′(π/3)=12tan(π/3)sec2(π/3)=183.(f^{-1})'(3)=\frac1{f'(\pi/3)} =\frac1{2\tan(\pi/3)\sec^2(\pi/3)} =\boxed{\frac1{8\sqrt3}}.(f−1)′(3)=f′(π/3)1=2tan(π/3)sec2(π/3)1=831.
