Question

求解一道大学高数的导数题？

试从dx/dy＝1/y'导出

Answer 1

dx/dy = 1/y'
d^2x/dy^2
=d/dy (1/y')
=d/dx (1/y') / (dy/dx)
=d/dx (1/y') / y'
= [-y''/(y')^2]/y'
=-y''/(y')^3
d^3x/dy^3
=d/dy [d^2x/dy^2]
=d/dx [d^2x/dy^2] / (dy/dx)
=d/dx [d^2x/dy^2] / y'
=d/dx [-y''/(y')^3] / y'
= { [-(y')^3. y''' + 3(y')^2. (y'')^2 ] /(y')^6 } /y'
= [-(y')^3. y''' + 3(y')^2. (y'')^2 ] /(y')^7
= [-y'. y''' + 3(y'')^2 ] /(y')^5

Answer 2

一种方法是看作1/e的x次方，套用指数函数的求导公式，结果是：1/e的x次方×ln(1/e)＝-(1/e的x次方)＝-e的-x次方
另一种方法是看作复合函数：y＝e的u次方，u＝-x，所以y的导数是：e的u次方×(-1)＝-e的-x次方