展开 4xy^2-2xy+((4x^2y-x^2-5)dy)/(dx)

解答

展开

4 x y^{2} - 2 x y + \frac{( 4 x ^{2} y - x ^{2} - 5 ) d y}{d x}

4 x y^{2} - 2 x y + \frac{4 x ^{2} y d y}{d x} - \frac{x ^{2} d y}{d x} - \frac{5 d y}{d x}

求解步骤

4 x y^{2} - 2 x y + \frac{( 4 x ^{2} y - x ^{2} - 5 ) d y}{d x}

乘开

(4 x^{2} y - x^{2} - 5) d y : 4 x^{2} y d y - x^{2} d y - 5 d y

= 4 x y^{2} - 2 x y + \frac{4 x ^{2} y d y - x ^{2} d y - 5 d y}{d x}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{4 x ^{2} y d y - x ^{2} d y - 5 d y}{d x} = \frac{4 x ^{2} y d y}{d x} - \frac{x ^{2} d y}{d x} - \frac{5 d y}{d x}

= 4 x y^{2} - 2 x y + \frac{4 x ^{2} y d y}{d x} - \frac{x ^{2} d y}{d x} - \frac{5 d y}{d x}

f a c t or x^{4} (δ (x - 2)) d x

e x p an d 3 \cdot (7 \cdot 4 y)

e x p an d \frac{- i ( 1 + i ) ^{n} - 1 + i + ( 1 + i ) ^{n}}{( 1 + i ) ^{n}}

e x p an d \frac{( 2 ( x + 2 ) ( n ^{\frac{1}{4}} ) )}{( n + 1 ) ^{\frac{1}{4}}}

e x p an d \frac{Z ^{2} \cdot K}{( Z - 0.95 ) \cdot ( Z + 0.95 )}