这一节介绍最常用的求微分定理，十分重要。证明就不证了，死记硬背也要记住这些定理。
Rule 1, Derivative of a constant function
If f has constant value f(x) = c, then
d f d x = d d x ( c ) = 0 \frac{df}{dx} = \frac{d}{dx}(c) = 0 d x df = d x d ( c ) = 0
Rule 2, Power rule for integers
if n is a positive integer, then
d d x x n = n x n − 1 \frac{d}{dx}x^n = nx^{n - 1} d x d x n = n x n − 1
if n is a negative integer and x ≠ 0 x \neq 0 x = 0
d d x x n = n x n − 1 \frac{d}{dx}x^n = nx^{n - 1} d x d x n = n x n − 1
Rule 3, Constant Multiple Rule
If u is a differentiable function of x, and c is a constant, then
d d x ( c u ) = c d u d x \frac{d}{dx}(cu) = c\frac{du}{dx} d x d ( c u ) = c d x d u
Rule 4, Dirivative of sum rule
If u and v are differentiable function of x, then their sum u + v is differentiable at every point where u and v are both differentiable. At such points,
d d x ( u + v ) = d u d x + d v d x \frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx} d x d ( u + v ) = d x d u + d x d v
Rule 4可以拓展到n个函数相加的情况，只要他们都是可微的。
Rule 5, Derivative product rule
If u and v are differentiable at x, then so is their product uv, and
d d x ( u v ) = u d v d x + v d u d x \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} d x d ( uv ) = u d x d v + v d x d u
Rule 6, Derivative quotient rule
If u an v are differentiable at x and if v ( x ) ≠ 0 v(x) \neq 0 v ( x ) = 0
d d x ( u v ) = v d u d x − u d v d x v 2 \frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} d x d ( v u ) = v 2 v d x d u − u d x d v
Second and higher-order derivatives
二阶至高阶导数，顾名思义，这里主要搞清楚约定的符号和英语里面对高阶导数的表达方式：
f ′ ′ ( x ) = d 2 y d x 2 = d d x ( d y d x ) = d y ′ d x = y ′ ′ = D 2 ( f ) ( x ) = D x 2 f ( x ) f''(x) = \frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{dy'}{dx} = y'' = D^2(f)(x) = D_x^2f(x) f ′′ ( x ) = d x 2 d 2 y = d x d ( d x d y ) = d x d y ′ = y ′′ = D 2 ( f ) ( x ) = D x 2 f ( x )
y ′ y' y ′ y ′ ′ y'' y ′′ d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y y ′ ′ ′ y''' y ′′′ y ( n ) y^{(n)} y ( n ) d n y d x n \frac{d^ny}{dx^n} d x n d n y D n D^n D n
练习
y = − x 2 + 3 y = -x^2 + 3 y = − x 2 + 3
d d x ( − x 2 + 3 ) = − 2 x \frac{d}{dx}(-x^2 + 3) = -2x d x d ( − x 2 + 3 ) = − 2 x
d d x ( − 2 x ) = − 2 \frac{d}{dx}(-2x) = -2 d x d ( − 2 x ) = − 2
y = x 2 + x + 8 y = x^2 + x + 8 y = x 2 + x + 8
y ′ = 2 x + 1 y' = 2x + 1 y ′ = 2 x + 1
y ′ ′ = 2 y'' = 2 y ′′ = 2
s = 5 t 3 − 3 t 5 s = 5t^3 - 3t^5 s = 5 t 3 − 3 t 5
s ′ = 15 t 2 − 15 t 4 s' = 15t^2 -15t^4 s ′ = 15 t 2 − 15 t 4
s ′ ′ = 30 t − 60 t 3 s'' = 30t - 60t^3 s ′′ = 30 t − 60 t 3
w = 3 z 7 − 7 z 3 + 21 z 2 w = 3z^7 - 7z^3 + 21z^2 w = 3 z 7 − 7 z 3 + 21 z 2
w ′ = 21 z 6 − 21 z 2 + 42 z w' = 21z^6 - 21z^2 + 42z w ′ = 21 z 6 − 21 z 2 + 42 z
w ′ ′ = 126 z 5 − 42 z + 42 w'' = 126z^5 - 42z + 42 w ′′ = 126 z 5 − 42 z + 42
y = 4 x 3 3 − x y = \frac{4x^3}{3} - x y = 3 4 x 3 − x
y ′ = 4 x 2 − 1 y' = 4x^2 - 1 y ′ = 4 x 2 − 1
y ′ ′ = 8 x y'' = 8x y ′′ = 8 x
y = x 3 3 + x 2 2 + x 4 y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{x}{4} y = 3 x 3 + 2 x 2 + 4 x
y ′ = x 2 + x + 1 4 y' = x^2 + x + \frac{1}{4} y ′ = x 2 + x + 4 1
y ′ ′ = 2 x + 1 y'' = 2x + 1 y ′′ = 2 x + 1
w = 3 z − 2 − 1 z w = 3z^{-2} - \frac{1}{z} w = 3 z − 2 − z 1
w ′ = − 6 z − 3 + 1 z 2 w' = -6z^{-3} + \frac{1}{z^2} w ′ = − 6 z − 3 + z 2 1
w ′ ′ = 18 z − 4 − 2 z 3 w'' = 18z^{-4} - \frac{2}{z^3} w ′′ = 18 z − 4 − z 3 2
s = − 2 t − 1 + 4 t 2 s = -2t^{-1} + \frac{4}{t^2} s = − 2 t − 1 + t 2 4
s ′ = 2 t − 2 − 8 t 3 s' = 2t^{-2} - \frac{8}{t^3} s ′ = 2 t − 2 − t 3 8
s ′ ′ = − 4 t − 3 + 24 t 4 s'' = -4t^{-3} + \frac{24}{t^4} s ′′ = − 4 t − 3 + t 4 24
y = 6 x 2 − 10 x − 5 x − 2 y = 6x^2 - 10x - 5x^{-2} y = 6 x 2 − 10 x − 5 x − 2
y ′ = 12 x − 10 + 10 x − 3 y' = 12x - 10 + 10x^{-3} y ′ = 12 x − 10 + 10 x − 3
y ′ ′ = 12 − 30 x − 4 y'' = 12 - 30x^{-4} y ′′ = 12 − 30 x − 4
y = 4 − 2 x − x − 3 y = 4 - 2x - x^{-3} y = 4 − 2 x − x − 3
y ′ = − 2 + 3 x − 4 y' = -2 + 3x^{-4} y ′ = − 2 + 3 x − 4
y ′ ′ = − 12 x − 5 y'' = -12x^{-5} y ′′ = − 12 x − 5
r = 1 3 s 2 − 5 2 s r = \frac{1}{3s^2} - \frac{5}{2s} r = 3 s 2 1 − 2 s 5
r ′ = − 2 3 s 3 + 5 2 s 2 r' = -\frac{2}{3s^3} + \frac{5}{2s^2} r ′ = − 3 s 3 2 + 2 s 2 5
r ′ ′ = 2 s 4 − 5 s 3 r'' = \frac{2}{s^4} - \frac{5}{s^3} r ′′ = s 4 2 − s 3 5
r = 12 ϕ − 4 ϕ 3 + 1 ϕ 4 r = \frac{12}{\phi} - \frac{4}{\phi^3} + \frac{1}{\phi^4} r = ϕ 12 − ϕ 3 4 + ϕ 4 1
r ′ = − 12 ϕ 2 + 12 ϕ 4 − 4 ϕ 5 r' = -\frac{12}{\phi^2} + \frac{12}{\phi^4} - \frac{4}{\phi^5} r ′ = − ϕ 2 12 + ϕ 4 12 − ϕ 5 4
r ′ ′ = 24 ϕ 3 − 48 ϕ 5 + 20 ϕ 6 r'' = \frac{24}{\phi^3} - \frac{48}{\phi^5} + \frac{20}{\phi^6} r ′′ = ϕ 3 24 − ϕ 5 48 + ϕ 6 20
y = ( 3 − x 2 ) ( x 3 − x + 1 ) y = (3 - x^2)(x^3 - x + 1) y = ( 3 − x 2 ) ( x 3 − x + 1 )
y ′ = − 2 x ( x 3 − x + 1 ) + ( 3 − x 2 ) ( 3 x 2 − 1 ) = − 5 x 4 + 12 x 2 − 2 x − 3 y' = -2x(x^3 - x + 1) + (3 - x^2)(3x^2 -1) = -5x^4 + 12x^2 -2x -3 y ′ = − 2 x ( x 3 − x + 1 ) + ( 3 − x 2 ) ( 3 x 2 − 1 ) = − 5 x 4 + 12 x 2 − 2 x − 3
y ′ = d d x ( − x 5 + 4 x 3 − x 2 − 3 x + 3 ) = − 5 x 4 + 12 x 2 − 2 x − 3 y' = \frac{d}{dx}(-x^5 + 4x^3 -x^2 -3x + 3) = -5x^4 + 12x^2 -2x -3 y ′ = d x d ( − x 5 + 4 x 3 − x 2 − 3 x + 3 ) = − 5 x 4 + 12 x 2 − 2 x − 3
y = ( x − 1 ) ( x 2 + x + 1 ) y = (x - 1)(x^2 + x + 1) y = ( x − 1 ) ( x 2 + x + 1 )
y ′ = x 2 + x + 1 + ( x − 1 ) ( 2 x + 1 ) = 3 x 2 y' = x^2 + x + 1 + (x - 1)(2x + 1) = 3x^2 y ′ = x 2 + x + 1 + ( x − 1 ) ( 2 x + 1 ) = 3 x 2
y ′ = d d x ( x 3 + x 2 + x − x 2 − x − 1 ) = 3 x 2 y' = \frac{d}{dx}(x^3 + x^2 + x - x^2 - x - 1) = 3x^2 y ′ = d x d ( x 3 + x 2 + x − x 2 − x − 1 ) = 3 x 2
y = ( x 2 + 1 ) ( x + 5 + 1 x ) y = (x^2 + 1)(x + 5 + \frac{1}{x}) y = ( x 2 + 1 ) ( x + 5 + x 1 )
y ′ = 2 x ( x + 5 + 1 x ) + ( x 2 + 1 ) ( 1 − 1 x 2 ) = 3 x 2 + 10 x + 2 − 1 x 2 y' = 2x(x + 5 + \frac{1}{x}) + (x^2 + 1)(1 - \frac{1}{x^2}) = 3x^2 + 10x + 2 - \frac{1}{x^2} y ′ = 2 x ( x + 5 + x 1 ) + ( x 2 + 1 ) ( 1 − x 2 1 ) = 3 x 2 + 10 x + 2 − x 2 1
y ′ = d d x ( x 3 + 5 x 2 + x + x + 5 + 1 x ) = 3 x 2 + 10 x + 2 − 1 x 2 y' = \frac{d}{dx}(x^3 + 5x^2 + x + x + 5 + \frac{1}{x}) = 3x^2 + 10x + 2 - \frac{1}{x^2} y ′ = d x d ( x 3 + 5 x 2 + x + x + 5 + x 1 ) = 3 x 2 + 10 x + 2 − x 2 1
y = ( x + 1 x ) ( x − 1 x + 1 ) y = (x + \frac{1}{x})(x - \frac{1}{x} + 1) y = ( x + x 1 ) ( x − x 1 + 1 )
y ′ = ( 1 − 1 x 2 ) ( x − 1 x + 1 ) + ( x + 1 x ) ( 1 + 1 x 2 ) = 2 x − 1 x 2 + 2 x 3 + 1 y' = (1 - \frac{1}{x^2})(x - \frac{1}{x} + 1) + (x + \frac{1}{x})(1 + \frac{1}{x^2}) = 2x - \frac{1}{x^2} + \frac{2}{x^3} + 1 y ′ = ( 1 − x 2 1 ) ( x − x 1 + 1 ) + ( x + x 1 ) ( 1 + x 2 1 ) = 2 x − x 2 1 + x 3 2 + 1
y ′ = d d x ( x 2 − 1 + x + 1 − 1 x 2 + 1 x ) = 2 x + 1 + 2 x 3 − 1 x 2 y' = \frac{d}{dx}(x^2 - 1 + x + 1 - \frac{1}{x^2} + \frac{1}{x}) = 2x + 1 + \frac{2}{x^3} - \frac{1}{x^2} y ′ = d x d ( x 2 − 1 + x + 1 − x 2 1 + x 1 ) = 2 x + 1 + x 3 2 − x 2 1
y = 2 x + 5 3 x − 2 y = \frac{2x + 5}{3x - 2} y = 3 x − 2 2 x + 5
y ′ = 2 ( 3 x − 2 ) − 3 ( 2 x + 5 ) ) ( 3 x − 2 ) 2 = − 19 9 x 2 − 12 x + 4 y' = \frac{2(3x - 2) - 3(2x + 5))}{(3x - 2)^2} = \frac{-19}{9x^2 - 12x + 4} y ′ = ( 3 x − 2 ) 2 2 ( 3 x − 2 ) − 3 ( 2 x + 5 )) = 9 x 2 − 12 x + 4 − 19
z = 2 x + 1 x 2 − 1 z = \frac{2x + 1}{x^2 - 1} z = x 2 − 1 2 x + 1
z ′ = 2 ( x 2 − 1 ) − 2 x ( 2 x + 1 ) ( x 2 − 1 ) 2 = − 2 x 2 − 2 x − 2 x 4 − 2 x 2 + 1 z' = \frac{2(x^2 - 1) - 2x(2x + 1)}{(x^2 - 1)^2} = \frac{-2x^2 - 2x - 2}{x^4 - 2x^2 + 1} z ′ = ( x 2 − 1 ) 2 2 ( x 2 − 1 ) − 2 x ( 2 x + 1 ) = x 4 − 2 x 2 + 1 − 2 x 2 − 2 x − 2
g ( x ) = x 2 − 4 x + 0.5 g(x) = \frac{x^2 - 4}{x + 0.5} g ( x ) = x + 0.5 x 2 − 4
g ′ ( x ) = 2 x ( x + 0.5 ) − x 2 + 4 ( x + 0.5 ) 2 = x 2 + x + 4 x 2 + 0.25 + x g'(x) = \frac{2x(x + 0.5) - x^2 + 4}{(x + 0.5)^2} = \frac{x^2 + x + 4}{x^2 + 0.25 + x} g ′ ( x ) = ( x + 0.5 ) 2 2 x ( x + 0.5 ) − x 2 + 4 = x 2 + 0.25 + x x 2 + x + 4
f ( t ) = t 2 − 1 t 2 + t − 2 f(t) = \frac{t^2 - 1}{t^2 + t - 2} f ( t ) = t 2 + t − 2 t 2 − 1
f ′ ( t ) = 2 t ( t 2 + t − 2 ) − ( 2 t + 1 ) ( t 2 − 1 ) ( t 2 + t − 2 ) 2 = t 2 − 2 t + 1 t 4 + 2 t 3 − 3 t 2 − 4 t + 4 f'(t) = \frac{2t(t^2 + t - 2) - (2t + 1)(t^2 - 1)}{(t^2 + t - 2)^2} = \frac{t^2 - 2t + 1}{t^4 + 2t^3 -3t^2 - 4t + 4} f ′ ( t ) = ( t 2 + t − 2 ) 2 2 t ( t 2 + t − 2 ) − ( 2 t + 1 ) ( t 2 − 1 ) = t 4 + 2 t 3 − 3 t 2 − 4 t + 4 t 2 − 2 t + 1
v = ( 1 − t ) ( 1 + t 2 ) − 1 v = (1 - t)(1 + t^2)^{-1} v = ( 1 − t ) ( 1 + t 2 ) − 1
v ′ = − ( 1 + t 2 ) − 2 t ( 1 − t ) ( 1 + t 2 ) 2 = t 2 − 2 t − 1 1 + t 4 + 2 t 2 v' = \frac{-(1 + t^2) - 2t(1 - t)}{(1 + t^2)^2} = \frac{t^2 -2t - 1}{1 + t^4 + 2t^2} v ′ = ( 1 + t 2 ) 2 − ( 1 + t 2 ) − 2 t ( 1 − t ) = 1 + t 4 + 2 t 2 t 2 − 2 t − 1
w = ( 2 x − 7 ) − 1 ( x + 5 ) w = (2x - 7)^{-1}(x + 5) w = ( 2 x − 7 ) − 1 ( x + 5 )
w ′ = 2 x − 7 − 2 x − 10 ( 2 x − 7 ) 2 = − 17 4 x 2 + 47 − 28 x w' = \frac{2x - 7 - 2x - 10}{(2x - 7)^2} = \frac{-17}{4x^2 + 47 - 28x} w ′ = ( 2 x − 7 ) 2 2 x − 7 − 2 x − 10 = 4 x 2 + 47 − 28 x − 17
f ( s ) = s − 1 s + 1 f(s) = \frac{\sqrt{s} - 1}{\sqrt{s} + 1} f ( s ) = s + 1 s − 1
f ′ ( s ) = 1 2 s ( s + 1 ) − 1 2 s ( s − 1 ) ( s + 1 ) 2 = 1 2 s + s + s s f'(s) = \frac{\frac{1}{2\sqrt{s}}(\sqrt{s} + 1) - \frac{1}{2\sqrt{s}}(\sqrt{s} - 1)}{(\sqrt{s} + 1)^2} = \frac{1}{2s + \sqrt{s} + s\sqrt{s}} f ′ ( s ) = ( s + 1 ) 2 2 s 1 ( s + 1 ) − 2 s 1 ( s − 1 ) = 2 s + s + s s 1
u = 5 x + 1 2 x u = \frac{5x + 1}{2\sqrt{x}} u = 2 x 5 x + 1
u ′ = 10 x − 1 x ( 5 x + 1 ) 4 x = 5 x − 1 4 x x u' = \frac{10\sqrt{x} - \frac{1}{\sqrt{x}}(5x + 1)}{4x} = \frac{5x - 1}{4x\sqrt{x}} u ′ = 4 x 10 x − x 1 ( 5 x + 1 ) = 4 x x 5 x − 1
v = 1 + x − 4 x x v = \frac{1 + x - 4\sqrt{x}}{x} v = x 1 + x − 4 x
v ′ = x ( 1 − 2 x ) − 1 − x + 4 x x 2 = 2 x − x x 2 x v' = \frac{x(1 - \frac{2}{\sqrt{x}}) - 1 - x + 4\sqrt{x}}{x^2} = \frac{2x - \sqrt{x}}{x^2\sqrt{x}} v ′ = x 2 x ( 1 − x 2 ) − 1 − x + 4 x = x 2 x 2 x − x
r = 2 ( 1 ϕ + ϕ ) r = 2(\frac{1}{\sqrt{\phi}} + \sqrt{\phi}) r = 2 ( ϕ 1 + ϕ )
r ′ = 0 − 1 ϕ ϕ + 1 ϕ = ϕ − 1 ϕ ϕ r' = \frac{0 - \frac{1}{\sqrt{\phi}}}{\phi} + \frac{1}{\sqrt{\phi}} = \frac{\phi - 1}{\phi\sqrt{\phi}} r ′ = ϕ 0 − ϕ 1 + ϕ 1 = ϕ ϕ ϕ − 1
y = 1 ( x 2 − 1 ) ( x 2 + x + 1 ) y = \frac{1}{(x^2 - 1)(x^2 + x + 1)} y = ( x 2 − 1 ) ( x 2 + x + 1 ) 1
y ′ = d d x ( 1 x 4 + x 3 − x − 1 ) = − 4 x 3 − 3 x 2 + 1 ( x 4 + x 3 − x − 1 ) 2 y' = \frac{d}{dx}(\frac{1}{x^4 + x^3 - x - 1}) = \frac{-4x^3 - 3x^2 + 1}{(x^4 + x^3 - x - 1)^2} y ′ = d x d ( x 4 + x 3 − x − 1 1 ) = ( x 4 + x 3 − x − 1 ) 2 − 4 x 3 − 3 x 2 + 1
y = ( x + 1 ) ( x + 2 ) ( x − 1 ) ( x − 2 ) y = \frac{(x + 1)(x + 2)}{(x - 1)(x - 2)} y = ( x − 1 ) ( x − 2 ) ( x + 1 ) ( x + 2 )
y ′ = d d x ( x + 1 x − 1 ) ⋅ x + 2 x − 2 + d d x ( x + 2 x − 2 ) ⋅ x + 1 x − 1 = − 2 ( x − 1 ) 2 ⋅ x + 2 x − 2 + − 4 ( x − 2 ) 2 ⋅ x + 1 x − 1 = − 6 x 2 + 8 ( x − 1 ) 2 ( x − 2 ) 2 y' = \frac{d}{dx}(\frac{x + 1}{x - 1})\cdot \frac{x + 2}{x - 2} + \frac{d}{dx}(\frac{x + 2}{x - 2})\cdot \frac{x + 1}{x - 1} = \frac{-2}{(x - 1)^2}\cdot \frac{x + 2}{x - 2} + \frac{-4}{(x - 2)^2}\cdot \frac{x + 1}{x - 1} = \frac{-6x^2 + 8}{(x - 1)^2(x - 2)^2} y ′ = d x d ( x − 1 x + 1 ) ⋅ x − 2 x + 2 + d x d ( x − 2 x + 2 ) ⋅ x − 1 x + 1 = ( x − 1 ) 2 − 2 ⋅ x − 2 x + 2 + ( x − 2 ) 2 − 4 ⋅ x − 1 x + 1 = ( x − 1 ) 2 ( x − 2 ) 2 − 6 x 2 + 8
