Question

复合函数求导？

第七第八题

Answer 1

复合函数的求导公式

Answer 2

7. y＝e^[-sin²(1/x)]，
dy/dx＝e^[-sin²(1/x)]
·[-2sin(1/x)·cos(1/x)·(-1/x²)]
＝{[sin(2/x)]e^[-sin²(1/x)]}/x²；
8. y＝√(x+√x)，
y'＝{1/[2√(x+√x)]}·[1+1/(2√x)]
＝{1/[2√(x+√x)]}·[(1+2√x)/(2√x)]
＝(1+2√x)/{4√[x(x+√x)]} .

Answer 3

7. y'= [e^(-(sin(1/x))^2)]' = e^[-(sin(1/x))^2] * [-(sin(1/x))^2] '
= e^(-(sin(1/x))^2 * (-2(sin(1/x) )(sin(1/x))'= e^(-(sin(1/x))^2 * (-2(sin(1/x) )cos(1/x)(1/x)'
= e^(-(sin(1/x))^2 * (-2(sin(1/x) )cos(1/x)(-1/x^2)= e^(-(sin(1/x))^2 *sin2x/x^2
8. y'= [(x+x^(1/2))^(1/2)]' = 1/[2(x+x^(1/2))] *[x+x^(1/2)]' = 1/[2(x+x^(1/2))] *[1+1/(2x^(1/2))]

Answer 4

解如下图所示

Answer 5

方法如下图所示，
请认真查看，
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