derivative of ({Q}(x,y)(x^2+y(x,d)^2)dy(x,d))

解

\frac{d}{d x} ((Q (x, y)) (x^{2} + y (x, d) (x, d)^{2}) d y (x, d))

d (\frac{d}{d x} (Q (x, y)) (x^{2} + y (x, d) (x, d)^{2}) y (x, d) + ((2 x + (\frac{d}{d x} (y (x, d)) (x, d)^{2} + 2 (x, d) (, d \cdot 1) y (x, d))) y (x, d) + \frac{d}{d x} (y (x, d)) (x^{2} + y (x, d) (x, d)^{2})) Q (x, y))

解答ステップ

\frac{d}{d x} ((Q (x, y)) (x^{2} + y (x, d) (x, d)^{2}) d y (x, d))

d

を定数として扱う

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= d \frac{d}{d x} (Q (x, y) (x^{2} + y (x, d) (x, d)^{2}) y (x, d))

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = Q (x, y), g = (x^{2} + y (x, d) (x, d)^{2}) y (x, d)

= d (\frac{d}{d x} (Q (x, y)) (x^{2} + y (x, d) (x, d)^{2}) y (x, d) + \frac{d}{d x} ((x^{2} + y (x, d) (x, d)^{2}) y (x, d)) Q (x, y))

\frac{d}{d x} ((x^{2} + y (x, d) (x, d)^{2}) y (x, d)) = (2 x + (\frac{d}{d x} (y (x, d)) (x, d)^{2} + 2 (x, d) (, d \cdot 1) y (x, d))) y (x, d) + \frac{d}{d x} (y (x, d)) (x^{2} + y (x, d) (x, d)^{2})

= d (\frac{d}{d x} (Q (x, y)) (x^{2} + y (x, d) (x, d)^{2}) y (x, d) + ((2 x + (\frac{d}{d x} (y (x, d)) (x, d)^{2} + 2 (x, d) (, d \cdot 1) y (x, d))) y (x, d) + \frac{d}{d x} (y (x, d)) (x^{2} + y (x, d) (x, d)^{2})) Q (x, y))