f ′ ( x 0 ) = lim ⁡ Δ x → 0 Δ y Δ x = lim ⁡ Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x \begin{align} & f^\prime(x_0) = \lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x} = \lim_{\Delta x\rightarrow 0}\frac{f(x_0 + \Delta x)-f(x_0)}{\Delta x} \end{align} ​f′(x0​)=Δx→0lim​ΔxΔy​=Δx→0lim​Δxf(x0​+Δx)−f(x0​)​​​

可记作 y ′ ∣ x = x 0 或 d y d x ∣ x = x 0 或 d [ f ( x ) ] d x ∣ x = x 0 可记作 y^\prime |_{x = x_0}或\frac{dy}{dx} |_{x = x_0}或\frac{d[f(x)]}{dx} |_{x = x_0} 可记作y′∣x=x0​​或dxdy​∣x=x0​​或dxd[f(x)]​∣x=x0​​

lim ⁡ x → x 0 f ( x ) − f ( x 0 ) x − x 0 = f ′ ( x ) \lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x - x_0} = f^\prime(x) x→x0​lim​x−x0​f(x)−f(x0​)​=f′(x)

C ′ = 0 C^{\prime} = 0 C′=0

( x ) ′ = 1 (x)^{\prime} = 1 (x)′=1

( x n ) ′ = n x n − 1 (x^n)^{\prime} = nx^{n-1} (xn)′=nxn−1

( s i n x ) ′ = c o s x (sinx)^{\prime} = cosx (sinx)′=cosx

( c o s x ) ′ = − s i n x (cosx)^{\prime} = - sinx (cosx)′=−sinx

( t a n x ) ′ = s e c 2 x (tanx)^{\prime} = sec^2x (tanx)′=sec2x

( c o t x ) ′ = − c s c 2 x (cotx)^{\prime} = -csc^2x (cotx)′=−csc2x

( s e c x ) ′ = s e c x t a n x (secx)^{\prime} = secxtanx (secx)′=secxtanx

( c s c x ) ′ = − c s c x c o t x (cscx)^{\prime} = -cscxcotx (cscx)′=−cscxcotx

( x ) ′ = 1 2 x (\sqrt{x})^{\prime} = \frac{1}{2\sqrt{x}} (x ​)′=2x ​1​

( a x ) = a x I n a (a^x) = a^xIna (ax)=axIna

( e x ) ′ = e x (e^x)^{\prime} = e^x (ex)′=ex

( l o g a x ) ′ = 1 x I n a (log_ax)^{\prime} = \frac{1}{xIna} (loga​x)′=xIna1​

( I n x ) ′ = 1 x (Inx)^{\prime} = \frac{1}{x} (Inx)′=x1​

( a c r s i n x ) ′ = 1 1 − x 2 (acrsinx)^{\prime} = \frac{1}{\sqrt{1-x^2}} (acrsinx)′=1−x2 ​1​

( a r c c o s x ) ′ = − 1 1 − x 2 (arccosx)^{\prime} = -\frac{1}{\sqrt{1-x^2}} (arccosx)′=−1−x2 ​1​

( a r c t a n x ) ′ = 1 1 + x 2 (arctanx)^{\prime} = \frac{1}{1+x^2} (arctanx)′=1+x21​

( a r c c o t x ) ′ = − 1 1 + x 2 (arccotx)^{\prime} = -\frac{1}{1+x^2} (arccotx)′=−1+x21​

一阶求导 { x = s i n t y = c o s 2 t \left\{\begin{matrix} x = sint\\ y = cos2t \end{matrix}\right. {x=sinty=cos2t​

d y d x = d y d t / d x d t = − 2 s i n 2 t c o s t = − 4 s i n t \frac{dy}{dx} = \frac{dy}{dt}/\frac{dx}{dt} = \frac{-2sin2t}{cost} = -4sint dxdy​=dtdy​/dtdx​=cost−2sin2t​=−4sint 二阶求导 { x = t 2 2 y = 1 − 4 \left\{\begin{matrix} x = \frac{t^2}{2}\\ y = 1-4 \end{matrix}\right. {x=2t2​y=1−4​

d y d x = d y d t / d x d t = − 1 t \frac{dy}{dx} = \frac{dy}{dt}/\frac{dx}{dt} = -\frac{1}{t} dxdy​=dtdy​/dtdx​=−t1​

d 2 y d x 2 = d ( y x ′ ) d x = d ( y x ′ ) d t / d x d t = ( y x ′ ) t ′ ( y 对 x 求导后对 t 求导 ) x t ′ = 1 t 3 \frac{d^2y}{dx^2} = \frac{d(y_x^\prime)}{dx} = \frac{d(y_x^\prime)}{dt}/\frac{dx}{dt} = \frac{(y_x^\prime)_t^\prime(y对x求导后对t求导) }{x_t^\prime}= \frac{1}{t^3} dx2d2y​=dxd(yx′​)​=dtd(yx′​)​/dtdx​=xt′​(yx′​)t′​(y对x求导后对t求导)​=t31​

函数两边对x求导，但是也同时对y求导，保留 y ′ y' y′，例如：

y 2 − 2 x y + 9 = 0 y^2 - 2xy + 9 = 0 y2−2xy+9=0，两边对x求导得 2 y ⋅ y ′ − 2 y − 2 x y ′ = 0 2y \cdot y' - 2y - 2xy' = 0 2y⋅y′−2y−2xy′=0，则

y ′ = 2 y 2 y − 2 x = y y − x y' = \frac{2y}{2y-2x} = \frac{y}{y-x} y′=2y−2x2y​=y−xy​

存在函数 f ( x , y ) f(x,y) f(x,y)，对函数求x的偏导数则将y视为常数对函数进行偏导，对函数求y的偏导同理。

已知 z = ( x 2 + y 2 ) e x + y ，求 ∂ z ∂ x 、 ∂ z ∂ y 、 ∂ 2 z ∂ x 2 、 ∂ 2 z ∂ y 2 、 ∂ 2 z ∂ x ∂ y 、 ∂ 2 z ∂ y ∂ x 已知z = (x^2 + y^2)e^{x+y}，求\frac{\partial z}{\partial x}、\frac{\partial z}{\partial y}、\frac{\partial^2 z}{\partial x^2}、\frac{\partial^2 z}{\partial y^2}、\frac{\partial^2 z}{\partial x\partial y}、\frac{\partial^2 z}{\partial y\partial x} 已知z=(x2+y2)ex+y，求∂x∂z​、∂y∂z​、∂x2∂2z​、∂y2∂2z​、∂x∂y∂2z​、∂y∂x∂2z​

∂ z ∂ x ( 函数 z 对 x 求偏导 ) = 2 x e x + y + ( x 2 + y 2 ) e x + y = ( x 2 + y 2 + 2 x ) e x + y \frac{\partial z}{\partial x}(函数z对x求偏导) = 2xe^{x+y} + (x^2 + y^2)e^{x+y} = (x^2 + y^2 + 2x)e^{x+y} ∂x∂z​(函数z对x求偏导)=2xex+y+(x2+y2)ex+y=(x2+y2+2x)ex+y

∂ z ∂ y ( 函数 z 对 y 求偏导 ) = 2 y e x + y + ( x 2 + y 2 ) e x + y = ( x 2 + y 2 + 2 y ) e x + y \frac{\partial z}{\partial y}(函数z对y求偏导) = 2ye^{x+y} + (x^2 + y^2)e^{x+y} = (x^2 + y^2 + 2y)e^{x+y} ∂y∂z​(函数z对y求偏导)=2yex+y+(x2+y2)ex+y=(x2+y2+2y)ex+y

∂ 2 z ∂ x 2 ( 函数 z 对 x 求二次偏导 ) = ( 2 x + 2 ) e x + y + ( x 2 + y 2 + 2 x ) e x + y = ( x 2 + y 2 + 4 x + 2 ) e x + y \frac{\partial^2 z}{\partial x^2}(函数z对x求二次偏导) = (2x+2)e^{x+y} + (x^2 + y^2 + 2x)e^{x+y} = (x^2 + y^2 + 4x + 2)e^{x+y} ∂x2∂2z​(函数z对x求二次偏导)=(2x+2)ex+y+(x2+y2+2x)ex+y=(x2+y2+4x+2)ex+y

∂ 2 z ∂ y 2 ( 函数 z 对 y 求二次偏导 ) = ( 2 y + 2 ) e x + y + ( x 2 + y 2 + 2 y ) e x + y = ( x 2 + y 2 + 4 y + 2 ) e x + y \frac{\partial^2 z}{\partial y^2}(函数z对y求二次偏导) = (2y+2)e^{x+y} + (x^2 + y^2 + 2y)e^{x+y} = (x^2 + y^2 + 4y + 2)e^{x+y} ∂y2∂2z​(函数z对y求二次偏导)=(2y+2)ex+y+(x2+y2+2y)ex+y=(x2+y2+4y+2)ex+y

∂ 2 z ∂ x ∂ y ( 函数 z 对 x 求偏导后对 y 求偏导 ) = 2 y e x + y + ( x 2 + y 2 + 2 x ) e x + y = ( x 2 + y 2 + 2 x + 2 y ) e x + y \frac{\partial^2 z}{\partial x\partial y}(函数z对x求偏导后对y求偏导) = 2ye^{x+y} + (x^2 + y^2 + 2x)e^{x+y} = (x^2 + y^2 + 2x + 2y)e^{x+y} ∂x∂y∂2z​(函数z对x求偏导后对y求偏导)=2yex+y+(x2+y2+2x)ex+y=(x2+y2+2x+2y)ex+y

∂ 2 z ∂ y ∂ x ( 函数 z 对 y 求偏导后对 x 求偏导 ) = 2 x e x + y + ( x 2 + y 2 + 2 y ) e x + y = ( x 2 + y 2 + 2 x + 2 y ) e x + y \frac{\partial^2 z}{\partial y\partial x}(函数z对y求偏导后对x求偏导) = 2xe^{x+y} + (x^2 + y^2 + 2y)e^{x+y} = (x^2 + y^2 + 2x + 2y)e^{x+y} ∂y∂x∂2z​(函数z对y求偏导后对x求偏导)=2xex+y+(x2+y2+2y)ex+y=(x2+y2+2x+2y)ex+y

求函数 f ( x , y ) = x 3 + y 3 − 3 x 2 − 3 y 2 f(x,y)=x^3 + y^3-3x^2-3y^2 f(x,y)=x3+y3−3x2−3y2 ∂ f ∂ x = 3 x 2 − 6 x \frac{\partial f}{\partial x} = 3x^2-6x ∂x∂f​=3x2−6x ∂ f ∂ x = 3 y 2 − 6 y \frac{\partial f}{\partial x} = 3y^2-6y ∂x∂f​=3y2−6y 令 ∂ f ∂ x = 0 ， ∂ f ∂ y = 0 令\frac{\partial f}{\partial x} = 0，\frac{\partial f}{\partial y} = 0 令∂x∂f​=0，∂y∂f​=0 得所有驻点为 ( 0 ， 0 ) ， ( 0 ， 2 ) ， ( 2 ， 0 ) ， ( 2 ， 2 ) (0，0)，(0，2)，(2，0)，(2，2) (0，0)，(0，2)，(2，0)，(2，2) 二阶偏导数为 ∂ 2 f ∂ x 2 = 6 x − 6 \frac{\partial ^2f}{\partial x^2} = 6x-6 ∂x2∂2f​=6x−6 ∂ 2 f ∂ x ∂ y = 0 \frac{\partial ^2f}{\partial x\partial y} = 0 ∂x∂y∂2f​=0 ∂ 2 f ∂ y 2 = 6 y − 6 \frac{\partial ^2f}{\partial y^2} = 6y-6 ∂y2∂2f​=6y−6

综上，极大值为 f ( 0 ， 0 ) = 0 f(0，0)=0 f(0，0)=0，极小值为 f ( 2 ， 2 ) = − 8 f(2，2)=-8 f(2，2)=−8

使用数学归纳法，计算出 f ′ ( x ) f^\prime(x) f′(x)、 f ′ ′ ( x ) f^{\prime\prime}(x) f′′(x)、 f ′ ′ ′ ( x ) f^{\prime\prime\prime}(x) f′′′(x)，然后总结结果得出 f ( n ) ( x ) f^{(n)}(x) f(n)(x)

将式子恒等变形分解、化简后使用基本公式计算：

( s i n x ) ( n ) = s i n ( x + n 2 π ) (sinx)^{(n)} = sin(x + \frac{n}{2}\pi) (sinx)(n)=sin(x+2n​π)

( s i n k x ) ( n ) = k n s i n ( k x + n 2 π ) (sinkx)^{(n)} = k^n sin(kx + \frac{n}{2}\pi) (sinkx)(n)=knsin(kx+2n​π)

( c o s x ) ( n ) = c o s ( x + n 2 π ) (cosx)^{(n)} = cos(x + \frac{n}{2}\pi) (cosx)(n)=cos(x+2n​π)

( c o s k x ) ( n ) = k n c o s ( k x + n 2 π ) (coskx)^{(n)} = k^n cos(kx + \frac{n}{2}\pi) (coskx)(n)=kncos(kx+2n​π)

( 1 a + b x ) ( n ) = ( − 1 ) n ⋅ b n n ! ⋅ 1 ( a x + b ) n + 1 (\frac{1}{a+bx})^{(n)} = (-1)^n \cdot b^n n! \cdot \frac{1}{(ax+b)^{n+1}} (a+bx1​)(n)=(−1)n⋅bnn!⋅(ax+b)n+11​

[ l n ( a x + b ) ] ( n ) = ( − 1 ) n − 1 ⋅ a n ( n − 1 ) ! ⋅ 1 ( a x + b ) n [ln(ax + b)]^{(n)} = (-1)^{n-1} \cdot a^n (n-1)! \cdot \frac{1}{(ax+b)^n} [ln(ax+b)](n)=(−1)n−1⋅an(n−1)!⋅(ax+b)n1​

[ u ( x ) v ( x ) ] ( n ) = ∑ k = 0 n C n k [ u ( x ) ] ( k ) [ v ( x ) ] ( n − k ) [u(x)v(x)]^{(n)} = \sum^n_{k=0}C_n^k [u(x)]^{(k)} [v(x)]^{(n-k)} [u(x)v(x)](n)=k=0∑n​Cnk​[u(x)](k)[v(x)](n−k) 其中 [ u ( x ) ] ( 0 ) = u ( x ) [u(x)]^{(0)} = u(x) [u(x)](0)=u(x)， [ v ( x ) ] ( 0 ) = v ( x ) [v(x)]^{(0)} = v(x) [v(x)](0)=v(x)

由两个函数乘积构成的函数，有一个函数为次数较低的多项式函数，由于阶数高于该次数的导数均为0，即 ( x n ) ( n ) = n ! (x^n)^{(n)} = n! (xn)(n)=n!， ( x n ) ( n + 1 ) = 0 (x^n)^{(n+1)} = 0 (xn)(n+1)=0，这时用该公式比较方便

用泰勒公式求f(x)在点 x 0 x_0 x0​的n阶导数 f ( n ) ( x 0 ) f^{(n)}(x_0) f(n)(x0​)

设 f ( x ) f(x) f(x)是定义在 ( − ∞ , + ∞ ) (-\infty, +\infty) (−∞,+∞)上的函数