Question

z=y+(L-x)(dy/dx),对x求导,怎么得的dz/dx=(dy/dx)-(dy/dx)+(L-x)(d^2y/d

Answer 1

dz/dx=dy/dx+d(L-x)/dx*(dy/dx)+(L-x)d(dy/dx)

=dy/dx-dy/dx+(L-x)d^2y/dx^2

d^2y/dx^2 是二阶导数 记住就行 !