第1次、第2次偏導関数の全て(Zx,Zy,Zxx,Zxy,Zyx,Zyy)を求めよ。 z=e^(xy

第1次、第2次偏導関数の全て(Zx,Zy,Zxx,Zxy,Zyx,Zyy)を求めよ。
z=e^(xy)sin(x^2＋2y^2)

No.2 回答者： ありものがたり 回答日時： 2020/08/02 12:05

すげぇ面倒くさい。

sinθ = (-i/2){ e^(iθ) - e^(-iθ) } を使って
z = e^(xy) (-i/2){ e^(i(x^2+2y^2)) - e^(-i(x^2+2y^2)) }
　= (-i/2){ e^(xy+i(x^2+2y^2)) - e^(xy-i(x^2+2y^2)) }
と変形したら、多少は楽かな？

(-2/i)Zx = (∂/∂x){ e^(xy+i(x^2+2y^2)) - e^(xy-i(x^2+2y^2)) }
　= (∂/∂x)e^(xy+i(x^2+2y^2)) - (∂/∂x)e^(xy-i(x^2+2y^2))
　= e^(xy+i(x^2+2y^2)) (∂/∂x)(xy+i(x^2+2y^2)) - e^(xy-i(x^2+2y^2)) (∂/∂x)(xy-i(x^2+2y^2))
　= e^(xy+i(x^2+2y^2)) (y+2ix) - e^(xy-i(x^2+2y^2)) (y-2ix)
　= e^(xy) { cos(x^2+2y^2) + i sin(x^2+2y^2) } (y+2ix) - e^(xy) { cos(x^2+2y^2) - i sin(x^2+2y^2) } (y-2ix)
　= (2i) e^(xy) { y sin(x^2+2y^2) + 2x cos(x^2+2y^2) },

(-2/i)Zy = (∂/∂y){ e^(xy+i(x^2+2y^2)) - e^(xy-i(x^2+2y^2)) }
　= (∂/∂y)e^(xy+i(x^2+2y^2)) - (∂/∂y)e^(xy-i(x^2+2y^2))
　= e^(xy+i(x^2+2y^2)) (∂/∂y)(xy+i(x^2+2y^2)) - e^(xy-i(x^2+2y^2)) (∂/∂y)(xy-i(x^2+2y^2))
　= e^(xy+i(x^2+2y^2)) (x+4iy) - e^(xy-i(x^2+2y^2)) (x-4iy)
　= e^(xy) { cos(x^2+2y^2) + i sin(x^2+2y^2) } (x+4iy) - e^(xy) { cos(x^2+2y^2) - i sin(x^2+2y^2) } (x-4iy)
　= (2i) e^(xy) { x sin(x^2+2y^2) + 4y cos(x^2+2y^2) },

(-2/i)Zxx = (∂/∂x){ e^(xy+i(x^2+2y^2)) (y+2ix) - e^(xy-i(x^2+2y^2)) (y-2ix) }
　= (∂/∂x){ e^(xy+i(x^2+2y^2)) (y+2ix) } - (∂/∂x){ e^(xy-i(x^2+2y^2)) (y-2ix) }
　= { e^(xy+i(x^2+2y^2)) (y+2ix)^2 + e^(xy+i(x^2+2y^2)) (2i) } - { e^(xy-i(x^2+2y^2)) (y-2ix)^2 + e^(xy-i(x^2+2y^2)) (-2i) }
　= e^(xy+i(x^2+2y^2)) (4x^2+4ixy-y^2+2i) - e^(xy-i(x^2+2y^2)) (4x^2-4ixy-y^2-2i)
　= e^(xy) { cos(x^2+2y^2) + i sin(x^2+2y^2) } (4x^2+4ixy-y^2+2i) - e^(xy) { cos(x^2+2y^2) - i sin(x^2+2y^2) } (4x^2-4ixy-y^2-2i)
　= (2i) e^(xy) { (4x^2-y^2) sin(x^2+2y^2) + (4xy+2) cos(x^2+2y^2) },

(-2/i)Zxy = (∂/∂y){ e^(xy+i(x^2+2y^2)) (y+2ix) - e^(xy-i(x^2+2y^2)) (y-2ix) }
　= (∂/∂y){ e^(xy+i(x^2+2y^2)) (y+2ix) } - (∂/∂y){ e^(xy-i(x^2+2y^2)) (y-2ix) }
　= { e^(xy+i(x^2+2y^2)) (x+4iy) (y+2ix) + e^(xy+i(x^2+2y^2)) (1) } - { e^(xy-i(x^2+2y^2)) (x-4iy) (y-2ix) + e^(xy-i(x^2+2y^2)) (1) }
　= e^(xy+i(x^2+2y^2)) (2ix^2-7xy+4iy^2+1) - e^(xy-i(x^2+2y^2)) (-2ix^2-7xy-4iy^2+1)
　= e^(xy) { cos(x^2+2y^2) + i sin(x^2+2y^2) } (2ix^2-7xy+4iy^2+1) - e^(xy) { cos(x^2+2y^2) + i sin(x^2+2y^2) } (-2ix^2-7xy-4iy^2+1)
　= (2i) e^(xy) { (-7xy+1) sin(x^2+2y^2) + (2x^2+4y^2) cos(x^2+2y^2) },

Zyx = Zxy,　；　Zxy が連続だから

(-2/i)Zyy = (∂/∂y){ e^(xy+i(x^2+2y^2)) (x+4iy) - e^(xy-i(x^2+2y^2)) (x-4iy) }
= (∂/∂y){ e^(xy+i(x^2+2y^2)) (x+4iy) } - (∂/∂y){ e^(xy-i(x^2+2y^2)) (x-4iy) }
= { e^(xy+i(x^2+2y^2)) (x+4iy)^2 + e^(xy+i(x^2+2y^2)) (4i) } - { e^(xy-i(x^2+2y^2)) (x-4iy)^2 + e^(xy-i(x^2+2y^2)) (-4i) }
= e^(xy+i(x^2+2y^2)) (x^2+8ixy-16y^2+4i) - e^(xy-i(x^2+2y^2)) (x^2-8ixy-16y^2-4i)
= e^(xy) { cos(x^2+2y^2) + i sin(x^2+2y^2) } (x^2+8ixy-16y^2+4i) - e^(xy) { cos(x^2+2y^2) - i sin(x^2+2y^2) } (x^2-8ixy-16y^2-4i)
= (2i) e^(xy) { (x^2-16y^2) sin(x^2+2y^2) + (8xy+4) cos(x^2+2y^2) }.

...なんだか、息切れがする

この回答へのお礼

丁寧な解説ありがとうございます

お礼日時：2020/08/02 13:11

No.1 回答者： varietyknowledge 回答日時： 2020/08/02 01:48

偏微分なのだから、対象の変数以外は定数とみなして微分すれば良い。

微分の内容としては、積の微分と合成関数の微分になる。