Question

求导复合函数x-a/√x²-2ax

Answer 1

y=(x-a)/√x²-2ax
那么对y 求导得到
y'= [(x-a)' *√x²-2ax -(x-a) *(√x²-2ax)'] / (x²-2ax)
显然
(x-a)'=1
(√x²-2ax)'= (2x-2a)/ 2√x²-2ax =(x-a) / √x²-2ax
于是得到
y'=[√x²-2ax -(x-a)* (x-a) / √x²-2ax] / (x²-2ax)
=[x²-2ax -(x-a)²] / (x²-2ax)^(3/2)
= -a² / (x²-2ax)^(3/2)

Answer 2

y=(x-a)/√x²-2ax
那么对y 求导得到
y'= [(x-a)' *√x²-2ax -(x-a) *(√x²-2ax)'] / (x²-2ax)
显然
(x-a)'=1
(√x²-2ax)'= (2x-2a)/ 2√x²-2ax =(x-a) / √x²-2ax
于是得到
y'=[√x²-2ax -(x-a)* (x-a) / √x²-2ax] / (x²-2ax)
=[x²-2ax -(x-a)²] / (x²-2ax)^(3/2)
= -a² / (x²-2ax)^(3/2)