f(x)=ln(x+√1+x^2) 求导
问题描述：
f(x)=ln(x+√1+x^2) 求导
答
f'(x)=[1+2x(x^2+1)^(-1/2)]/x+√(1+x^2)
答
f(x)=ln(x+√1+x^2)
f'(x)=1/(x+√(1+x^2) *(x+√1+x^2)'
=1/(x+√(1+x^2)*(1+(√1+x^2)'
=1/(x+√(1+x^2)*(1+1/2*√(x^2+1) *(x^2)')
=1/(x+√(1+x^2)*(1+1/2*√(x^2+1) *2x)
=(1+x/√(x^2+1))/(x+√(1+x^2) 分子分母乘x-√(1+x^2)
=(1+x/√(x^2+1)(x-√(x^2+1)/(x^2-1-x^2)
=(1+x/√(x^2+1)(√(x^2+1)-x)
=√(x^2+1)-x+x-x^2/√(x^2+1)
=√(x^2+1-x^2√(x^2+1)/(x^2+1)
=√(x^2+1)(1-x^2/(x^2+1))
=√(x^2+1)(x^2+1-x^2)/(x^2+1)
=√(x^2+1)/(x^2+1)