(\partial)/(\partial y)(2x+y+1)

解答

\frac{\partial}{\partial y} (2 x + y + 1)

1

求解步骤

\frac{\partial}{\partial y} (2 x + y + 1)

将

x

视为常数

使用微分加减法定则:

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

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

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

\frac{\partial}{\partial y} (y) = 1

\frac{\partial}{\partial y} (1) = 0

= 0 + 1 + 0

化简

= 1

d er i v a t i v e \frac{x ^{2}}{6 + x}

x \to - 1 lim (- x^{2} + 4 x - 5)

\frac{d}{d x} (\frac{2 x ^{2}}{( x ^{2} + x ) ^{3}})

t an g e n t f (x) = 4 x^{2} - 2 x + 4, at x = 0

t an g e n t f (x) = - 2 x^{4} + 5 x^{2} - 3, (1, 0)