(\partial)/(\partial x)((y)(x^3y^3-x^3+1))

解答

\frac{\partial}{\partial x} ((y) (x^{3} y^{3} - x^{3} + 1))

y (3 y^{3} x^{2} - 3 x^{2})

求解步骤

\frac{\partial}{\partial x} ((y) (x^{3} y^{3} - x^{3} + 1))

将

y

视为常数

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= y \frac{\partial}{\partial x} (x^{3} y^{3} - x^{3} + 1)

使用微分加减法定则:

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

= y (\frac{\partial}{\partial x} (x^{3} y^{3}) - \frac{\partial}{\partial x} (x^{3}) + \frac{\partial}{\partial x} (1))

\frac{\partial}{\partial x} (x^{3} y^{3}) = 3 y^{3} x^{2}

\frac{\partial}{\partial x} (x^{3}) = 3 x^{2}

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

= y (3 y^{3} x^{2} - 3 x^{2} + 0)

化简

= y (3 y^{3} x^{2} - 3 x^{2})

e x p an d x - (- 3)

s im pl i f y \frac{e ^{ia x}}{( z + b ) ^{2} ( z - b ) ^{2}}

e x p an d \frac{1}{( ( x - r ) ^{2} + y ^{2} + z ^{2} ) ^{\frac{5}{2}}}

e x p an d \frac{1}{9 + 6 x}

e x p an d 1 + sin (b) (12, - 9)