设函数y=ln(x-2),则dy⼀dx=多少

2025-05-22 13:20:06
推荐回答(2个)
回答1:

dy=d(ln(x-2))
=(1/(x-2))*d(x-2)
=(1/(x-2))dx
所以,dy/dx=1/(x-2)

回答2:

设z=x-2
dy/dx=dz/dx×dz/dy=1×1/y=1/(x-2)