Question

这道题怎么解

Answer 1

y=ln√ (x+(sinx)^2)
=(1/2)ln(x+(sinx)^2)
dy/dx
= (1/2)[1/(x+(sinx)^2)] .d/dx(x+(sinx)^2)
= (1/2)[1/(x+(sinx)^2)] . (1+2sinx.cosx)
=(1+2sinx.cosx)/ [ 2(x+(sinx)^2)]

Answer 2

y=ln√[x+(sinx)^2]的导数
y′=1/√[x+(sinx)^2]*1/{2√[x+(sinx)^2]} *(1+2sinx cosx)
=(sinx+cosx)^2/{2[x+(sinx)^2]}

Answer 3

如