derivative of a/(1+\frac{b{x+c}})

解答

\frac{d}{d x} (\frac{a}{1 + \frac{b}{x + c}})

\frac{ab}{( x + c + b ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{a}{1 + \frac{b}{x + c}})

将常数提出:

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

= a \frac{d}{d x} (\frac{1}{1 + \frac{b}{x + c}})

使用指数法则:

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

= a \frac{d}{d x} ((1 + \frac{b}{x + c})^{- 1})

使用连锁律:

- \frac{1}{( 1 + \frac{b}{x + c} ) ^{2}} \frac{d}{d x} (1 + \frac{b}{x + c})

= - \frac{1}{( 1 + \frac{b}{x + c} ) ^{2}} \frac{d}{d x} (1 + \frac{b}{x + c})

\frac{d}{d x} (1 + \frac{b}{x + c}) = - \frac{b}{( x + c ) ^{2}}

= a (- \frac{1}{( 1 + \frac{b}{x + c} ) ^{2}} (- \frac{b}{( x + c ) ^{2}}))

化简

a (- \frac{1}{( 1 + \frac{b}{x + c} ) ^{2}} (- \frac{b}{( x + c ) ^{2}})) : \frac{ab}{( x + c + b ) ^{2}}

= \frac{ab}{( x + c + b ) ^{2}}

\int 6 x^{2} d x

x \to 0 lim (x - \frac{1}{ln ∣ x ∣})

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 6

x \to - 2 - lim (\frac{x}{x ^{2} - 4})

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