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第二节 导数的计算

Answer

I. 导数公式

(C)=0(C)' = 0(xμ)=μxμ1(x^{\mu}) = \mu x^{\mu - 1}
(sinx)=cosx(\sin x)' = \cos x(cosx)=sinx(\cos x)' = -\sin x
(tanx)=sec2x(\tan x)' = \sec^2 x(cotx)=csc2x(\cot x)' = -\csc^2 x
(secx)=secxtanx(\sec x)' = \sec x \tan x(cscx)=cscxcotx(\csc x)' = -\csc x \cot x
(ax)=axlna,(a>0,a1)(a^x) = a^x \ln a ,\, (a \gt 0, a \neq 1)(ex)=ex(e^x)' = e^x
(logax)=1xlna(a>0,a1)(\log_a x)' = \dfrac{1}{x \ln a} \, (a \gt 0, a \neq 1)(lnx)=1x(\ln x)' = \dfrac{1}{x}
(arcsinx)=11x2(\arcsin x)' = \dfrac{1}{\sqrt{1 - x^2}}(arccosx)=11x2(\arccos x)' = -\dfrac{1}{\sqrt{1 - x^2}}
(arctanx)=11+x2(\arctan x)' = \dfrac{1}{1 + x^2}(arccotx)=11+x2(\mathrm{arccot} \, x)' = -\dfrac{1}{1 + x^2}

II. 导数的四则运算法则

  • 例1:求下列函数的导函数:

    • (1)y=xarcsinx2+4x2+ln2y = x \arcsin \frac{x}{2} + \sqrt{4 - x^2} + \ln 2

      Answer
    • (2)y=etan1xcos1xy = e^{\tan \frac{1}{x}} \cos \frac{1}{x}

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    • (3)y=esin1x+1+x1xexy = e^{\sin \frac{1}{x}} + \frac{1 + x}{1 - x} e^{\sqrt{x}}

      Answer
    • (4)y=xaa+axa+aaxy = x^{a^a} + a^{x^a} + a^{a^x}

      Answer

III. 反函数的求导法则

y=f(x)y = f(x)可导,且f(x)0f(x) \neq 0,则其反函数x=φ(y)x = \varphi(y)的导数为:

  • 例2:设y=f(x)y = f(x)具有连续的一阶导数,且其反函数为x=φ(y)x = \varphi(y),若f(1)=2f(1) = 2f(1)=3f'(1) = 3,则φ(2)\varphi'(2)的值为何?

    Answer

    由反函数的导数的求导法则可知:

    又因为:

    则:

    所以当y=2y = 2时对应的x=1x = 1

IV. 复合函数的求导法则

链式求导法则:设y=f(u)y = f(u)u=φ(x)u = \varphi(x)都可导,则y=f[φ(x)]y = f\left[ \varphi(x) \right]可导,且dydx=dydududx\dfrac{\mathrm{d} y}{\mathrm{d} x} = \dfrac{\mathrm{d} y}{\mathrm{d} u} \cdot \dfrac{\mathrm{d} u}{\mathrm{d} x},即:

  • 例3:求下列函数的导函数:

    • (1)y=earctanx+1x1y = e^{\arctan \frac{x + 1}{x - 1}}

      Answer
    • (2)y=ln2tan(x2+1)y = \ln^2 \tan (x^2 + 1)

      Answer
    • (3)y=sin[f(x+1x1)]y = \sin \left[ f(\sqrt{\frac{x + 1}{x - 1}}) \right]

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  • 例4:已知y=f(3x23x+2)y = f(\dfrac{3x - 2}{3x + 2})f(x)=arctanx2f'(x) = \arctan x^2,则dydxx=0\dfrac{\mathrm{d} y}{\mathrm{d} x}|_{x = 0}为何值?

    Answer

    根据链式法则求得dydx\dfrac{\mathrm{d} y}{\mathrm{d} x}

    然后将x=0x = 0代入:

V. 隐函数的求导

y=f(x)y = f(x)由方程F(x,y)=0F(x, y) = 0确定,在F(x,y)=0F(x, y) = 0中视yyxx的函数,两边同时对xx求导,即可得dydx\dfrac{\mathrm{d} y}{\mathrm{d} x}

  • 例5:函数y=y(x)y = y(x)是由方程xy+ex+y=2x+1xy + e^{x + y} = 2x + 1所确定的隐函数,则dydxx=0\dfrac{\mathrm{d} y}{\mathrm{d} x}|_{x = 0}为何值?

    Answer

    两边同时对xx求导:

    然后将x=0x = 0代入原方程求出yy的值:

    最后代入x=0x = 0y=0y = 0yy'的表达式中:

    即可得dydxx=0=1\dfrac{\mathrm{d} y}{\mathrm{d} x}|_{x = 0} = 1

VI. 幂指函数的求导

对于幂指函数y=u(x)v(x)(u(x)>0,u(x)1)y = u(x)^{v(x)} \, (u(x) \gt 0, u(x) \neq 1),若u(x)u(x)v(x)v(x)均可导,常用求导方式有两种:

  1. 转换为指数恒等式u(x)v(x)=ev(x)lnu(x)u(x)^{v(x)} = e^{v(x) \ln u(x)},再利用复合函数的求导法则对其求导

  2. 两边取对数得lny=v(x)ln[u(x)]\ln y = v(x) \ln \left[ u(x) \right],再对其两边关于xx求导

最终解得:

  • 例6:设y=(1+x2)xy = (1 + x^2)^{\sqrt{x}},求yy'.

    Answer

VII. 由参数方程所确立的函数的求导

设函数y=y(x)y = y(x)由以下参数方程确立:

x(t)x(t)y(t)y(t)皆可导,且x(t)0x'(t) \neq 0,则dydx=dydtdtdx=y(t)x(t)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \cdot \dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{y'(t)}{x'(t)}

  • 例7:设函数y=y(x)y = y(x)

    确立,求dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}.

    Answer