高等数学A1 周维祺
(f+g)′(x)=f′(x)+g′(x)(f+g)'(x)=f'(x)+g'(x) (f+g)′(x)=f′(x)+g′(x)
(f+g)′(x)=limh→0(f+g)(x+h)−(f+g)(x)h=limh→0f(x+h)−f(x)h+g(x+h)−g(x)h=f′(x)+g′(x)\begin{align*}(f+g)'(x)&=\lim_{h\to0}\frac{(f+g)(x+h)-(f+g)(x)}{h} \\ &=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h} \\ &=f'(x)+g'(x) \end{align*}(f+g)′(x)=h→0limh(f+g)(x+h)−(f+g)(x)=h→0limhf(x+h)−f(x)+hg(x+h)−g(x)=f′(x)+g′(x)
(fg)′(x)=f′(x)g(x)+g′(x)f(x)(fg)'(x)=f'(x)g(x)+g'(x)f(x) (fg)′(x)=f′(x)g(x)+g′(x)f(x)
(fg)′(x)=limh→0(fg)(x+h)−(fg)(x)h=limh→0f(x+h)g(x+h)−f(x+h)g(x)+f(x+h)g(x)−f(x)g(x)h=limh→0f(x+h)g(x+h)−g(x)h+g(x)f(x+h)−f(x)h=f(x)g′(x)+g(x)f′(x)\begin{align*}(fg)'(x)&=\lim_{h\to0}\frac{(fg)(x+h)-(fg)(x)}{h} \\ &=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h} \\ &=\lim_{h\to0}f(x+h)\frac{g(x+h)-g(x)}{h}+g(x)\frac{f(x+h)-f(x)}{h} \\ &=f(x)g'(x)+g(x)f'(x) \end{align*}(fg)′(x)=h→0limh(fg)(x+h)−(fg)(x)=h→0limhf(x+h)g(x+h)−f(x+h)g(x)+f(x+h)g(x)−f(x)g(x)=h→0limf(x+h)hg(x+h)−g(x)+g(x)hf(x+h)−f(x)=f(x)g′(x)+g(x)f′(x)
(fg)(n)=∑k=0n(nk)f(n−k)g(k)(fg)^{(n)}=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}g^{(k)} (fg)(n)=k=0∑n(kn)f(n−k)g(k)
(f(x)g(x))′=f′(x)g(x)−g′(x)f(x)(g(x))2,g(x)≠0\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-g'(x)f(x)}{\left(g(x)\right)^2},\quad g(x)\neq0 (g(x)f(x))′=(g(x))2f′(x)g(x)−g′(x)f(x),g(x)=0
(1g(x))′=limh→01g(x+h)−1g(x)h=limh→0g(x)−g(x+h)h⋅1g(x)g(x+h)=−g′(x)(g(x))2\begin{align*}\left(\frac{1}{g(x)}\right)'&=\lim_{h\to0}\frac{\frac{1}{g(x+h)}-\frac{1}{g(x)}}{h} \\ &=\lim_{h\to0}\frac{g(x)-g(x+h)}{h}\cdot\frac{1}{g(x)g(x+h)} \\ &=-\frac{g'(x)}{\left(g(x)\right)^2} \end{align*}(g(x)1)′=h→0limhg(x+h)1−g(x)1=h→0limhg(x)−g(x+h)⋅g(x)g(x+h)1=−(g(x))2g′(x)
完成证明的剩余步骤
计算y=tanxy=\tan xy=tanx的导数
计算y=2x3sinx+lnxy=2x^3\sin x+\ln xy=2x3sinx+lnx的导数
求导可以与加法和数乘交换顺序
(af(x))′=af′(x)\left(af(x)\right)'=af'(x) (af(x))′=af′(x)
(f[g(x)])′=f′[g(x)]g′(x)\left(f[g(x)]\right)'=f'[g(x)]g'(x) (f[g(x)])′=f′[g(x)]g′(x)
(f[g(x)])′=limh→0f[g(x+h)]−f[g(x)]h=limh→0f[g(x+h)]−f[g(x)]g(x+h)−g(x)⋅g(x+h)−g(x)h=f′[g(x)]g′(x)\begin{align*}\left(f[g(x)]\right)'&=\lim_{h\to0}\frac{f[g(x+h)]-f[g(x)]}{h} \\ &=\lim_{h\to0}\frac{f[g(x+h)]-f[g(x)]}{g(x+h)-g(x)}\cdot\frac{g(x+h)-g(x)}{h} \\ &=f'[g(x)]g'(x) \end{align*}(f[g(x)])′=h→0limhf[g(x+h)]−f[g(x)]=h→0limg(x+h)−g(x)f[g(x+h)]−f[g(x)]⋅hg(x+h)−g(x)=f′[g(x)]g′(x)
(令u=g(x),u+Δu=g(x+h)u=g(x),\quad u+\Delta u=g(x+h)u=g(x),u+Δu=g(x+h))
计算下列函数的导数
y=1−x2y=\sqrt{1-x^2}y=1−x2
y=ln(cosx2)y=\ln(\cos x^2)y=ln(cosx2)
若x=f(y)x=f(y)x=f(y)有反函数y=f−1(x)y=f^{-1}(x)y=f−1(x)，且f,f−1f,f^{-1}f,f−1都可导，则
(f−1(x))′=1f′(y)=1f′(f−1(x))\left(f^{-1}(x)\right)'=\frac{1}{f'(y)}=\frac{1}{f'\left(f^{-1}(x)\right)} (f−1(x))′=f′(y)1=f′(f−1(x))1
(f−1(x))′=limh→0f−1(x+h)−f−1(x)x+h−x=limΔy→0y+Δy−yf(y+Δy)−f(y)=1f′(y)\begin{align*}\left(f^{-1}(x)\right)'&=\lim_{h\to0}\frac{f^{-1}(x+h)-f^{-1}(x)}{x+h-x} \\ &=\lim_{\Delta y\to0}\frac{y+\Delta y-y}{f(y+\Delta y)-f(y)} \\ &=\frac{1}{f'(y)} \\ \end{align*} (f−1(x))′=h→0limx+h−xf−1(x+h)−f−1(x)=Δy→0limf(y+Δy)−f(y)y+Δy−y=f′(y)1
y=axy=a^xy=ax (a>0,a≠1)(a>0, a\neq1)(a>0,a=1)
y=xxy=x^xy=xx
f(x)={x2sin1xx<00x=0x2cos1xx>0f(x)=\begin{dcases}x^2\sin\frac{1}{x} & x<0 \\ 0 & x=0 \\ x^2\cos\frac{1}{x} & x>0\end{dcases} f(x)=⎩⎨⎧x2sinx10x2cosx1x<0x=0x>0
（1）计算x≠0x\neq0x=0时f(x)f(x)f(x)的导数
（2）利用定义验证f(x)f(x)f(x)在x=0x=0x=0处是否可导
（3）f′(x)f'(x)f′(x)是否连续？