Question

设函数f(x)=arccot2x-tan4x,则f′(0)等于多少?

Answer 1

(arctanx)'=1/1+x^2

(arccotx)'=-1/1+x^2

f'(x)=-2/(1+4x^2)-4sec^2(4x)

f'(0)=-2-4=-6

(g(f(x))'

=d(g(f(x))/d(f(x) * d(f(x)/dx

如:y=(x^2-1)^2

y'=d((x^2-1)^2)/d(x^2-1) * d(x^2-1)/dx

y'=2(x^2-1)*2x