Question

f(x)=ln(x+√1+x^2) 求导

Answer 1

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)