(\partial)/(\partial x)(x^2+y^2-2x-2y)

解答

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

2 x - 2

求解步骤

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

将

y

视为常数

使用微分加减法定则:

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

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

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (y^{2}) = 0

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

\frac{\partial}{\partial x} (2 y) = 0

= 2 x + 0 - 2 - 0

化简

= 2 x - 2

a re a f (x) = 2 sin (x) + sin (2 x), y = 0, 0 \leq x \leq π

x^{^{'}} = x - 15

\frac{d ^{3}}{d x ^{3}} (\frac{4}{x})

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

d er i v a t i v e f (x) = \frac{4}{( x ^{2} + 1 ) ^{2}}