y=tan(x+y) 求二阶导数

如题所述

y'= dy/dx =sec^2(x+y)·(1+y');
→[sec^2(x+y) -1]·y'=sec^2(x+y);
→[tan^2(x+y) ]·y'=sec^2(x+y);
→y'=1/sin^2(x+y);

则:
y'' =dy' /dx
=d[sin^(-2)(x+y)] /dx
=(-2)·sin^(-3)(x+y) ·cos(x+y)·(1+y')
=-2·sin^(-3)(x+y) ·cos(x+y)·[1+sin^(-2)(x+y)]
=-2·cos(x+y)·[sin^(-3)(x+y) +sin^(-5)(x+y)]
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第1个回答  2016-04-18
y=tan(x-y)两边求导
y' = (1-y')/cos²(x-y)
y' = 1/[1+cos²(x-y)]
再次求导
y" = -2cos(x-y)[-sin(x-y)](1-y')/[1+cos²(x-y)]²
= sin(2x-2y)(1-y')/[1+cos²(x-y)]²
= sin(2x-2y){1 - 1/[1+cos²(x-y)]}/[1+cos²(x-y)]²
= sin(2x-2y)cos²(x-y)/[1+cos²(x-y)]³本回答被网友采纳

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