设z=ln(x-y2),则∂z/∂y=()
设z=ln(x-y2),则∂z/∂y=()
A、1/(x-y2)
B、-2y/(x-y2)
C、(1-2y)/(x-y2)
D、(x-2y)/(x-y2)
【正确答案】:B
【题目解析】:考查的知识点是复合函数求导的链式法则.令u=x-y2,则z=lnu,∂z/∂y=dz/du•∂u/∂y=1/u•∂u/∂y=-2y/(1-y)2,故答案选B.
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