Question

一道微积分题目求解

如图 f(x)=arctan(1-x/1+x)求f（x）的n阶导在0处的值

Answer 1

f'(x)=1/[1+(1-x)^2/(1+x)^2]*-2/(1+x^2)]
=-2/[(1+x)^2+(1-x)^2]
=1/(1+x^2),
f''历缓滑(x)=2x/(1+x^2)^2,
f'''(x)=2/(1+x^2)^2-8x^2/(1+x^2)^3
=(2-6x^2)/(1+x^2)^3,
f(4)(x)=-12x/(1+x^2)^3-6x(2-6x^2)/(1+x^2)^4
=(-12x-12x^3-12x+36x^3)/(1+x^2)^4
=(-12x+24x^3)/(1+x^2)^4,
…哪空…
仅供参考。肢腊