derivative of (x+y^2+(x+1)^2)

解

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

4 x + 2 y \frac{d y}{d x} + 2 (x \frac{d y}{d x} + y) + 2

解答ステップ

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

簡素化

(x + y)^{2} + (x + 1)^{2} : 2 x^{2} + 2 x y + 2 x + y^{2} + 1

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

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (2 x^{2}) = 4 x

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

\frac{d}{d x} (2 x) = 2

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

\frac{d}{d x} (1) = 0

= 4 x + 2 (y + x \frac{d y}{d x}) + 2 + 2 y \frac{d y}{d x} + 0

簡素化

= 4 x + 2 y \frac{d y}{d x} + 2 (x \frac{d y}{d x} + y) + 2