(\partial)/(\partial x)(-(x-4)^2-(y-1)^2)

解答

\frac{\partial}{\partial x} (- (x - 4)^{2} - (y - 1)^{2})

- 2 x + 8

求解步骤

\frac{\partial}{\partial x} (- (x - 4)^{2} - (y - 1)^{2})

化简

- (x - 4)^{2} - (y - 1)^{2} : - x^{2} + 8 x + 2 y - y^{2} - 17

= \frac{\partial}{\partial x} (- x^{2} + 8 x + 2 y - y^{2} - 17)

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} (8 x) + \frac{\partial}{\partial x} (2 y) - \frac{\partial}{\partial x} (y^{2}) - \frac{\partial}{\partial x} (17)

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

\frac{\partial}{\partial x} (8 x) = 8

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

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

\frac{\partial}{\partial x} (17) = 0

= - 2 x + 8 + 0 - 0 - 0

化简

= - 2 x + 8

\int \frac{1}{81 + 16 x ^{2}} d x

x \to 0 lim (- \frac{3}{1 - 3 x})

d er i v a t i v e f (x) = cot^{2} (sin (θ))

\frac{d}{d x} (sec (x) y)

\frac{d y}{d t} = y - 5