(\partial)/(\partial y)((x+y)^{e^{y^2}})

解

\frac{\partial}{\partial y} ((x + y)^{e^{y^{2}}})

(2 e^{y^{2}} y ln (x + y) + \frac{e ^{y^{2}}}{x + y}) (x + y)^{e^{y^{2}}}

解答ステップ

\frac{\partial}{\partial y} ((x + y)^{e^{y^{2}}})

x

を定数として扱う

指数の規則を適用する:

a^{b} = e^{b l n (a)}

(x + y)^{e^{y^{2}}} = e^{e^{y^{2}} l n (x + y)}

= \frac{\partial}{\partial y} (e^{e^{y^{2}} l n (x + y)})

連鎖法則を適用:

e^{e^{y^{2}} l n (x + y)} \frac{\partial}{\partial y} (e^{y^{2}} ln (x + y))

= e^{e^{y^{2}} l n (x + y)} \frac{\partial}{\partial y} (e^{y^{2}} ln (x + y))

\frac{\partial}{\partial y} (e^{y^{2}} ln (x + y)) = 2 e^{y^{2}} y ln (x + y) + \frac{e ^{y^{2}}}{x + y}

= e^{e^{y^{2}} l n (x + y)} (2 e^{y^{2}} y ln (x + y) + \frac{e ^{y^{2}}}{x + y})

e^{e^{y^{2}} l n (x + y)} = (x + y)^{e^{y^{2}}}

= (2 e^{y^{2}} y ln (x + y) + \frac{e ^{y^{2}}}{x + y}) (x + y)^{e^{y^{2}}}