導(dǎo)數(shù)答題題目及答案解析1.題目：求函數(shù)\(f(x)=x^3-3x^2+2\)在\(x=1\)處的導(dǎo)數(shù)。答案解析：首先，我們需要找到函數(shù)\(f(x)\)的導(dǎo)數(shù)\(f'(x)\)。根據(jù)導(dǎo)數(shù)的定義，我們對(duì)\(f(x)\)求導(dǎo)得到：\[f'(x)=\fracz3jilz61osys{dx}(x^3-3x^2+2)=3x^2-6x\]然后，我們將\(x=1\)代入\(f'(x)\)中，計(jì)算得到：\[f'(1)=3(1)^2-6(1)=3-6=-3\]所以，函數(shù)\(f(x)\)在\(x=1\)處的導(dǎo)數(shù)是\(-3\)。2.題目：計(jì)算函數(shù)\(g(x)=\sin(x)\)的導(dǎo)數(shù)。答案解析：根據(jù)三角函數(shù)的導(dǎo)數(shù)公式，我們知道\(\sin(x)\)的導(dǎo)數(shù)是\(\cos(x)\)。因此，函數(shù)\(g(x)\)的導(dǎo)數(shù)為：\[g'(x)=\cos(x)\]3.題目：給定函數(shù)\(h(x)=e^x\)，求其在\(x=0\)處的導(dǎo)數(shù)值。答案解析：指數(shù)函數(shù)\(e^x\)的導(dǎo)數(shù)是其本身，即\(\fracz3jilz61osys{dx}(e^x)=e^x\)。因此，函數(shù)\(h(x)\)的導(dǎo)數(shù)為：\[h'(x)=e^x\]將\(x=0\)代入\(h'(x)\)中，我們得到：\[h'(0)=e^0=1\]所以，函數(shù)\(h(x)\)在\(x=0\)處的導(dǎo)數(shù)值是\(1\)。4.題目：求函數(shù)\(k(x)=\ln(x)\)的導(dǎo)數(shù)。答案解析：自然對(duì)數(shù)函數(shù)\(\ln(x)\)的導(dǎo)數(shù)是\(\frac{1}{x}\)。因此，函數(shù)\(k(x)\)的導(dǎo)數(shù)為：\[k'(x)=\frac{1}{x}\]5.題目：計(jì)算函數(shù)\(m(x)=\sqrt{x}\)的導(dǎo)數(shù)。答案解析：函數(shù)\(m(x)=\sqrt{x}\)可以寫作\(m(x)=x^{1/2}\)。根據(jù)冪函數(shù)的導(dǎo)數(shù)規(guī)則，我們有：\[m'(x)=\fracz3jilz61osys{dx}(x^{1/2})=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}\]6.題目：求函數(shù)\(n(x)=\frac{1}{x}\)的導(dǎo)數(shù)。答案解析：函數(shù)\(n(x)=\frac{1}{x}\)可以寫作\(n(x)=x^{-1}\)。根據(jù)冪函數(shù)的導(dǎo)數(shù)規(guī)則，我們得到：\[n'(x)=\fracz3jilz61osys{dx}(x^{-1})=-1x^{-2}=-\frac{1}{x^2}\]7.題目：計(jì)算函數(shù)\(p(x)=x^2-4x+3\)的導(dǎo)數(shù)，并求在\(x=2\)處的導(dǎo)數(shù)值。答案解析：首先，我們對(duì)函數(shù)\(p(x)\)求導(dǎo)：\[p'(x)=\fracz3jilz61osys{dx}(x^2-4x+3)=2x-4\]然后，我們將\(x=2\)代入\(p'(x)\)中，計(jì)算得到：\[p'(2)=2(2)-4=4-4=0\]所以，函數(shù)\(p(x)\)在\(x=2\)處的導(dǎo)數(shù)值是\(0\)。8.題目：求函數(shù)\(q(x)=\tan(x)\)的導(dǎo)數(shù)。答案解析：根據(jù)三角函數(shù)的導(dǎo)數(shù)公式，我們知道\(\tan(x)\)的導(dǎo)數(shù)是\(\sec^2(x)\)。因此，函數(shù)\(q(x)\)的導(dǎo)數(shù)為：\[q'(x)=\sec^2(x)\]9.題目：計(jì)算函數(shù)\(r(x)=\arcsin(x)\)的導(dǎo)數(shù)。答案解析：反正弦函數(shù)\(\arcsin(x)\)的導(dǎo)數(shù)是\(\frac{1}{\sqrt{1-x^2}}\)。因此，函數(shù)\(r(x)\)的導(dǎo)數(shù)為：\[r'(x)=\frac{1}{\sqrt{1-x^2}}\]10.題目：求函數(shù)\(s(x)=\frac{x^2}{x+1}\)的導(dǎo)數(shù)。答案解析：這是一個(gè)商的導(dǎo)數(shù)問(wèn)題，我們可以使用商的導(dǎo)數(shù)法則，即\(\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\)。這里\(u=x^2\)和\(v=x+1\)，所以\(u'=2x\)和\(v'=1\)。應(yīng)用商的導(dǎo)數(shù)法則，我們得到：\[s'(x)=\frac{(2x)(x+1)-(x^2)(1)}{(x+1)^2}=\fra