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

解答

\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

求解步骤

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

化简

(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)

使用微分加减法定则:

(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

化简

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

in v erse l a pl a ce \frac{- 20}{s ( s + 10 )}

\int_{1}^{3} \frac{2}{t ^{3}} d t

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

\int a - x^{2} d x

n = 1 \sum \infty \frac{1}{3 n - 2}