第1个回答 2019-08-23
f(x)=x.ln[x+ √ (1+x^2) ]
f'(x)
=ln[x+ √ (1+x^2) ] + x . d/dx{ ln[x+ √ (1+x^2) ] }
=ln[x+ √ (1+x^2) ] + x . {1/ln[x+ √ (1+x^2) ] } . d/dx [x+ √ (1+x^2) ]
=ln[x+ √ (1+x^2) ] + x . {1/ln[x+ √ (1+x^2) ] } . [1+ x/√ (1+x^2) ]
=ln[x+ √ (1+x^2) ] + x .[ √ (1+x^2) +x] /{ ln[x+ √ (1+x^2) ] .√ (1+x^2) }