$\mathbb{R}^2$ 上の関数 $f(x, y) = x^2 y$ と $g(x, y) = (x+y)e^y$ の偏導関数をそれぞれ求める。解析学偏導関数多変数関数2025/6/301. 問題の内容R2\mathbb{R}^2R2 上の関数 f(x,y)=x2yf(x, y) = x^2 yf(x,y)=x2y と g(x,y)=(x+y)eyg(x, y) = (x+y)e^yg(x,y)=(x+y)ey の偏導関数をそれぞれ求める。2. 解き方の手順関数 f(x,y)=x2yf(x, y) = x^2 yf(x,y)=x2y の偏導関数を求める。fx=∂f∂x=∂(x2y)∂x=2xyf_x = \frac{\partial f}{\partial x} = \frac{\partial (x^2 y)}{\partial x} = 2xyfx=∂x∂f=∂x∂(x2y)=2xyfy=∂f∂y=∂(x2y)∂y=x2f_y = \frac{\partial f}{\partial y} = \frac{\partial (x^2 y)}{\partial y} = x^2fy=∂y∂f=∂y∂(x2y)=x2関数 g(x,y)=(x+y)eyg(x, y) = (x+y)e^yg(x,y)=(x+y)ey の偏導関数を求める。gx=∂g∂x=∂((x+y)ey)∂x=eyg_x = \frac{\partial g}{\partial x} = \frac{\partial ((x+y)e^y)}{\partial x} = e^ygx=∂x∂g=∂x∂((x+y)ey)=eygy=∂g∂y=∂((x+y)ey)∂y=∂(x+y)∂yey+(x+y)∂(ey)∂y=ey+(x+y)ey=(x+y+1)eyg_y = \frac{\partial g}{\partial y} = \frac{\partial ((x+y)e^y)}{\partial y} = \frac{\partial (x+y)}{\partial y} e^y + (x+y) \frac{\partial (e^y)}{\partial y} = e^y + (x+y)e^y = (x+y+1)e^ygy=∂y∂g=∂y∂((x+y)ey)=∂y∂(x+y)ey+(x+y)∂y∂(ey)=ey+(x+y)ey=(x+y+1)ey3. 最終的な答えf(x,y)=x2yf(x, y) = x^2 yf(x,y)=x2y の偏導関数は:fx=2xyf_x = 2xyfx=2xyfy=x2f_y = x^2fy=x2g(x,y)=(x+y)eyg(x, y) = (x+y)e^yg(x,y)=(x+y)ey の偏導関数は:gx=eyg_x = e^ygx=eygy=(x+y+1)eyg_y = (x+y+1)e^ygy=(x+y+1)ey