derivative of (-1/(2i+(y-x)j))

解答

\frac{d}{d x} (\frac{- 1}{2 i + ( y - x ) j})

\frac{j \frac{d y}{d x} - j}{( 2 i + j ( y - x ) ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{- 1}{2 i + ( y - x ) j})

化简

\frac{- 1}{2 i + ( y - x ) j} : - \frac{1}{2 i + j ( y - x )}

= \frac{d}{d x} (- \frac{1}{2 i + j ( y - x )})

将常数提出:

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

= - \frac{d}{d x} (\frac{1}{2 i + j ( y - x )})

使用指数法则:

\frac{1}{a} = a^{- 1}

= - \frac{d}{d x} ((2 i + j (y - x))^{- 1})

使用连锁律:

- \frac{1}{( 2 i + j ( y - x ) ) ^{2}} \frac{d}{d x} (2 i + j (y - x))

= - \frac{1}{( 2 i + j ( y - x ) ) ^{2}} \frac{d}{d x} (2 i + j (y - x))

\frac{d}{d x} (2 i + j (y - x)) = j \frac{d y}{d x} - j

= - (- \frac{1}{( 2 i + j ( y - x ) ) ^{2}} (j \frac{d y}{d x} - j))

化简

- (- \frac{1}{( 2 i + j ( y - x ) ) ^{2}} (j \frac{d y}{d x} - j)) : \frac{j \frac{d y}{d x} - j}{( 2 i + j ( y - x ) ) ^{2}}

= \frac{j \frac{d y}{d x} - j}{( 2 i + j ( y - x ) ) ^{2}}

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

\frac{d}{d t} (\frac{1}{2 \cdot a \cdot t ^{2}})

d er i v a t i v e y = \frac{1}{x ^{2} + x ^{2}}

d er i v a t i v e y = (x^{2} + 3 x)^{4}

d er i v a t i v e (1 + x^{3})^{- 1}