derivative of ((2y+3/(4x+5))^2)

解答

\frac{d}{d x} ((\frac{2 y + 3}{4 x + 5})^{2})

\frac{2 ( 2 y + 3 ) ( 2 \frac{d y}{d x} ( 4 x + 5 ) - 4 ( 2 y + 3 ) )}{( 4 x + 5 ) ^{3}}

求解步骤

\frac{d}{d x} ((\frac{2 y + 3}{4 x + 5})^{2})

使用连锁律:

2 \cdot \frac{2 y + 3}{4 x + 5} \frac{d}{d x} (\frac{2 y + 3}{4 x + 5})

= 2 \cdot \frac{2 y + 3}{4 x + 5} \frac{d}{d x} (\frac{2 y + 3}{4 x + 5})

\frac{d}{d x} (\frac{2 y + 3}{4 x + 5}) = \frac{2 \frac{d y}{d x} ( 4 x + 5 ) - 4 ( 2 y + 3 )}{( 4 x + 5 ) ^{2}}

= 2 \cdot \frac{2 y + 3}{4 x + 5} \cdot \frac{2 \frac{d y}{d x} ( 4 x + 5 ) - 4 ( 2 y + 3 )}{( 4 x + 5 ) ^{2}}

化简

2 \cdot \frac{2 y + 3}{4 x + 5} \cdot \frac{2 \frac{d y}{d x} ( 4 x + 5 ) - 4 ( 2 y + 3 )}{( 4 x + 5 ) ^{2}} : \frac{2 ( 2 y + 3 ) ( 2 \frac{d y}{d x} ( 4 x + 5 ) - 4 ( 2 y + 3 ) )}{( 4 x + 5 ) ^{3}}

= \frac{2 ( 2 y + 3 ) ( 2 \frac{d y}{d x} ( 4 x + 5 ) - 4 ( 2 y + 3 ) )}{( 4 x + 5 ) ^{3}}

t an g e n t f (x) = \frac{4 x}{x - 5}, (7, 14)

\frac{d u}{d x}

n = 0 \sum \infty n^{2} + n

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

\int_{0}^{1} \frac{x}{x + 1} d x